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How to Solve Logarithms

Last Updated: October 5, 2023 Fact Checked

This article was reviewed by Grace Imson, MA . Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 241,699 times.

Logarithms might be intimidating, but solving a logarithm is much simpler once you realize that logarithms are just another way to write out exponential equations. Once you rewrite the logarithm into a more familiar form, you should be able to solve it as you would solve any standard exponential equation.

Before You Begin: Learn to Express a Logarithmic Equation Exponentially [1] X Research source [2] X Research source

Step 1 Know the logarithm definition.

  • If and only if: b y = x
  • b does not equal 1
  • In the same equation, y is the exponent and x is the exponential expression that the logarithm is set equal to.

Step 2 Look at the equation.

  • Example: 1024 = ?

Step 4 Apply the exponent to the base.

  • This could also be written as: 4 5

Step 5 Rewrite your final answer.

  • Example: 4 5 = 1024

Method One: Solve for X

Step 1 Isolate the logarithm.

  • log 3 ( x + 5) + 6 - 6 = 10 - 6
  • log 3 ( x + 5) = 4

Step 2 Rewrite the equation in exponential form.

  • Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 4; b = 3; x = x + 5
  • Rewrite the equation so that: b y = x
  • 3 4 = x + 5

Step 3 Solve for x.

  • 3 * 3 * 3 * 3 = x + 5
  • 81 - 5 = x + 5 - 5

Step 4 Write your final answer.

  • Example: x = 76

Method Two: Solve for X Using the Logarithmic Product Rule [3] X Research source [4] X Research source

Step 1 Know the product rule.

  • log b (m * n) = log b (m) + log b (n)

Step 2 Isolate the logarithm to one side of the equation.

  • log 4 (x + 6) + log 4 (x) = 2 - log 4 (x) + log 4 (x)
  • log 4 (x + 6) + log 4 (x) = 2

Step 3 Apply the product rule.

  • log 4 [(x + 6) * x] = 2
  • log 4 (x 2 + 6x) = 2

Step 4 Rewrite the equation in exponential form.

  • Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 4 ; x = x 2 + 6x
  • 4 2 = x 2 + 6x

Step 5 Solve for x.

  • 4 * 4 = x 2 + 6x
  • 16 = x 2 + 6x
  • 16 - 16 = x 2 + 6x - 16
  • 0 = x 2 + 6x - 16
  • 0 = (x - 2) * (x + 8)
  • x = 2; x = -8

Step 6 Write your answer.

  • Example: x = 2
  • Note that you cannot have a negative solution for a logarithm, so you can discard x - 8 as a solution.

Method Three: Solve for X Using the Logarithmic Quotient Rule [5] X Research source

Step 1 Know the quotient rule.

  • log b (m / n) = log b (m) - log b (n)

Step 2 Isolate the logarithm to one side of the equation.

  • log 3 (x + 6) - log 3 (x - 2) = 2 + log 3 (x - 2) - log 3 (x - 2)
  • log 3 (x + 6) - log 3 (x - 2) = 2

Step 3 Apply the quotient rule.

  • log 3 [(x + 6) / (x - 2)] = 2

Step 4 Rewrite the equation in exponential form.

  • Comparing this equation to the definition [ y = log b (x) ], you can conclude that: y = 2; b = 3; x = (x + 6) / (x - 2)
  • 3 2 = (x + 6) / (x - 2)

Step 5 Solve for x.

  • 3 * 3 = (x + 6) / (x - 2)
  • 9 = (x + 6) / (x - 2)
  • 9 * (x - 2) = [(x + 6) / (x - 2)] * (x - 2)
  • 9x - 18 = x + 6
  • 9x - x - 18 + 18 = x - x + 6 + 18
  • 8x / 8 = 24 / 8

Step 6 Write your final answer.

  • Example: x = 3

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  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut43_logfun.htm#logdef
  • ↑ https://www.mathsisfun.com/algebra/logarithms.html
  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut46_logeq.htm
  • ↑ https://www.youtube.com/watch?v=fnhFneOz6n8
  • ↑ https://www.wtamu.edu/academic/anns/mps/math/mathlab/col_algebra/col_alg_tut44_logprop.htm

About This Article

Grace Imson, MA

To solve a logarithm, start by identifying the base, which is "b" in the equation, the exponent, which is "y," and the exponential expression, which is "x." Then, move the exponential expression to one side of the equation, and apply the exponent to the base by multiplying the base by itself the number of times indicated in the exponent. Finally, rewrite your final answer as an exponential expression. To learn how to solve for "x" in a logarithm, scroll down! Did this summary help you? Yes No

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Solving Logarithmic Equations

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Equations involving logarithms and unknown variables can often be solved by employing the definition of the logarithm, as well as several of its basic properties :

\( \log_x(a) + \log_x(b) = \log_x(ab) \)

\(\log_x(a) - \log_x(b) = \log_x\big(\frac ab\big) \)

\(a\log_x(b) = \log_x(b^a) \)

\(\log_x(a) = \frac{\log_y(a)}{\log_y(x)} = \frac1{\log_a(x)}\) for any positive real number \(y \)

\(x^{\log_x(a)} = a\).

Common mistakes to watch out for include

\(\log_x(a) \cdot \log_x(b) \neq \log_x(ab) \)

\(\frac{\log_x(a)}{\log_x(b)} \neq \log_x\big(\frac{a}{b}\big)\).

The general strategy is to consolidate the logarithms using these properties, and then to take both sides of the equation to the appropriate power in order to eliminate the logarithms if possible.

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Mathematics LibreTexts

4.7: Exponential and Logarithmic Equations

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  • Page ID 64854

Learning Objectives

  • Use like bases to solve exponential equations.
  • Use logarithms to solve exponential equations.
  • Use the definition of a logarithm to solve logarithmic equations.
  • Use the one-to-one property of logarithms to solve logarithmic equations.
  • Solve applied problems involving exponential and logarithmic equations.

In 1859, an Australian landowner named Thomas Austin released \(24\) rabbits into the wild for hunting. Because Australia had few predators and ample food, the rabbit population exploded. In fewer than ten years, the rabbit population numbered in the millions.

Seven rabbits in front of a brick building.

Uncontrolled population growth, as in the wild rabbits in Australia, can be modeled with exponential functions. Equations resulting from those exponential functions can be solved to analyze and make predictions about exponential growth. In this section, we will learn techniques for solving exponential functions.

Using Like Bases to Solve Exponential Equations

The first technique involves two functions with like bases. Recall that the one-to-one property of exponential functions tells us that, for any real numbers \(b\), \(S\), and \(T\), where \(b>0\), \(b≠1\), \(b^S=b^T\) if and only if \(S=T\).

In other words, when an exponential equation has the same base on each side, the exponents must be equal. This also applies when the exponents are algebraic expressions. Therefore, we can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then, we use the fact that exponential functions are one-to-one to set the exponents equal to one another, and solve for the unknown.

For example, consider the equation \(3^{4x−7}=\dfrac{3^{2x}}{3}\). To solve for \(x\), we use the division property of exponents to rewrite the right side so that both sides have the common base, \(3\). Then we apply the one-to-one property of exponents by setting the exponents equal to one another and solving for \(x\):

\[\begin{align*} 3^{4x-7}&= \dfrac{3^{2x}}{3}\\ 3^{4x-7}&= \dfrac{3^{2x}}{3^1} \qquad &&\text{Rewrite 3 as } 3^1\\ 3^{4x-7}&= 3^{2x-1} \qquad &&\text{Use the division property of exponents}\\ 4x-7&= 2x-1 \qquad &&\text{Apply the one-to-one property of exponents}\\ 2x&= 6 \qquad &&\text{Subtract 2x and add 7 to both sides}\\ x&= 3 \qquad &&\text{Divide by 3} \end{align*}\]

USING THE ONE-TO-ONE PROPERTY OF EXPONENTIAL FUNCTIONS TO SOLVE EXPONENTIAL EQUATIONS

For any algebraic expressions \(S\) and \(T\), and any positive real number \(b≠1\),

\[\begin{align} b^S=b^T\text{ if and only if } S=T \end{align}\]

How to: Given an exponential equation with the form \(b^S=b^T\), where \(S\) and \(T\) are algebraic expressions with an unknown, solve for the unknown.

  • Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form \(b^S=b^T\).
  • Use the one-to-one property to set the exponents equal.
  • Solve the resulting equation, \(S=T\), for the unknown.

Example \(\PageIndex{1}\): Solving an Exponential Equation with a Common Base

Solve \(2^{x−1}=2^{2x−4}\).

\[\begin{align*} 2^{x-1}&= 2^{2x-4} \qquad &&\text{The common base is 2}\\ x-1&= 2x-4 \qquad &&\text{By the one-to-one property the exponents must be equal}\\ x&= 3 \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{1}\)

Solve \(5^{2x}=5^{3x+2}\).

\(x=−2\)

Rewrite Equations So All Powers Have the Same Base

Sometimes the common base for an exponential equation is not explicitly shown. In these cases, we simply rewrite the terms in the equation as powers with a common base, and solve using the one-to-one property.

For example, consider the equation \(256=4^{x−5}\). We can rewrite both sides of this equation as a power of \(2\). Then we apply the rules of exponents, along with the one-to-one property, to solve for \(x\):

\[\begin{align*} 256&= 4^{x-5}\\ 2^8&= {(2^2)}^{x-5} \qquad &&\text{Rewrite each side as a power with base 2}\\ 2^8&= 2^{2x-10} \qquad &&\text{Use the one-to-one property of exponents}\\ 8&= 2x-10 \qquad &&\text{Apply the one-to-one property of exponents}\\ 18&= 2x \qquad &&\text{Add 10 to both sides}\\ x&= 9 \qquad &&\text{Divide by 2} \end{align*}\]

How to: Given an exponential equation with unlike bases, use the one-to-one property to solve it.

  • Rewrite each side in the equation as a power with a common base.

Example \(\PageIndex{2}\): Solving Equations by Rewriting Them to Have a Common Base

Solve \(8^{x+2}={16}^{x+1}\).

\[\begin{align*} 8^{x+2}&= {16}^{x+1}\\ {(2^3)}^{x+2}&= {(2^4)}^{x+1} \qquad &&\text{Write 8 and 16 as powers of 2}\\ 2^{3x+6}&= 2^{4x+4} \qquad &&\text{To take a power of a power, multiply exponents}\\ 3x+6&= 4x+4 \qquad &&\text{Use the one-to-one property to set the exponents equal}\\ x&= 2 \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{2}\)

Solve \(5^{2x}={25}^{3x+2}\).

\(x=−1\)

Example \(\PageIndex{3}\): Solving Equations by Rewriting Roots with Fractional Exponents to Have a Common Base

Solve \(2^{5x}=\sqrt{2}\).

\[\begin{align*} 2^{5x}&= 2^{\frac{1}{2}} \qquad &&\text{Write the square root of 2 as a power of 2}\\ 5x&= \dfrac{1}{2} \qquad &&\text{Use the one-to-one property}\\ x&= \dfrac{1}{10} \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{3}\)

Solve \(5^x=\sqrt{5}\).

\(x=\dfrac{1}{2}\)

Q&A: Do all exponential equations have a solution? If not, how can we tell if there is a solution during the problem-solving process?

No. Recall that the range of an exponential function is always positive. While solving the equation, we may obtain an expression that is undefined.

Example \(\PageIndex{4}\): Solving an Equation with Positive and Negative Powers

Solve \(3^{x+1}=−2\).

This equation has no solution. There is no real value of \(x\) that will make the equation a true statement because any power of a positive number is positive.

Figure \(\PageIndex{2}\) shows that the two graphs do not cross so the left side is never equal to the right side. Thus the equation has no solution.

Graph of 3^(x+1)=-2 and y=-2. The graph notes that they do not cross.

Exercise \(\PageIndex{4}\)

Solve \(2^x=−100\).

The equation has no solution.

Solving Exponential Equations Using Logarithms

Sometimes the terms of an exponential equation cannot be rewritten with a common base. In these cases, we solve by taking the logarithm of each side. Recall, since \(\log(a)=\log(b)\) is equivalent to \(a=b\), we may apply logarithms with the same base on both sides of an exponential equation.

How to: Given an exponential equation in which a common base cannot be found, solve for the unknown

  • If one of the terms in the equation has base 10, use the common logarithm.
  • If none of the terms in the equation has base 10, use the natural logarithm.
  • Use the rules of logarithms to solve for the unknown.

Example \(\PageIndex{5}\): Solving an Equation Containing Powers of Different Bases

Solve \(5^{x+2}=4^x\).

\[\begin{align*} 5^{x+2}&= 4^x \qquad &&\text{There is no easy way to get the powers to have the same base}\\ \ln5^{x+2}&= \ln4^x \qquad &&\text{Take ln of both sides}\\ (x+2)\ln5&= x\ln4 \qquad &&\text{Use laws of logs}\\ x\ln5+2\ln5&= x\ln4 \qquad &&\text{Use the distributive law}\\ x\ln5-x\ln4&= -2\ln5 \qquad &&\text{Get terms containing x on one side, terms without x on the other}\\ x(\ln5-\ln4)&= -2\ln5 \qquad &&\text{On the left hand side, factor out an x}\\ x\ln \left (\dfrac{5}{4} \right )&= \ln \left (\dfrac{1}{25} \right ) \qquad &&\text{Use the laws of logs}\\ x&=\dfrac{\ln \left (\dfrac{1}{25} \right )}{\ln \left (\dfrac{5}{4} \right )} \qquad &&\text{Divide by the coefficient of x} \end{align*}\]

Exercise \(\PageIndex{5}\)

Solve \(2^x=3^{x+1}\).

\(x=\dfrac{\ln3}{\ln \left (\dfrac{2}{3} \right )}\)

Q&A: Is there any way to solve \(2^x=3^x\)?

Yes. The solution is \(0\).

Equations Containing \(e\)

One common type of exponential equations are those with base \(e\). This constant occurs again and again in nature, in mathematics, in science, in engineering, and in finance. When we have an equation with a base \(e\) on either side, we can use the natural logarithm to solve it.

How to: Given an equation of the form \(y=Ae^{kt}\), solve for \(t\).

  • Divide both sides of the equation by \(A\).
  • Apply the natural logarithm of both sides of the equation.
  • Divide both sides of the equation by \(k\).

Example \(\PageIndex{6}\): Solve an Equation of the Form \(y = Ae^{kt}\)

Solve \(100=20e^{2t}\).

\[\begin{align*} 100&= 20e^{2t}\\ 5&= e^{2t} \qquad &&\text{Divide by the coefficient of the power}\\ \ln5&= 2t \qquad &&\text{Take ln of both sides. Use the fact that } ln(x) \text{ and } e^x \text{ are inverse functions}\\ t&= \dfrac{\ln5}{2} \qquad &&\text{Divide by the coefficient of t} \end{align*}\]

Using laws of logs, we can also write this answer in the form \(t=\ln\sqrt{5}\). If we want a decimal approximation of the answer, we use a calculator.

Exercise \(\PageIndex{6}\)

Solve \(3e^{0.5t}=11\).

\(t=2\ln \left (\dfrac{11}{3} \right )\) or \(\ln{ \left (\dfrac{11}{3} \right )}^2\)

Q&A: Does every equation of the form \(y=Ae^{kt}\) have a solution?

No. There is a solution when \(k≠0\),and when \(y\) and \(A\) are either both 0 or neither 0, and they have the same sign. An example of an equation with this form that has no solution is \(2=−3e^t\).

Example \(\PageIndex{7}\): Solving an Equation That Can Be Simplified to the Form \(y=Ae^{kt}\)

Solve \(4e^{2x}+5=12\).

\[\begin{align*} 4e^{2x}+5&= 12\\ 4e^{2x}&= 7 \qquad &&\text{Combine like terms}\\ e^{2x}&= \dfrac{7}{4} \qquad &&\text{Divide by the coefficient of the power}\\ 2x&= \ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Take ln of both sides}\\ x&= \dfrac{1}{2}\ln \left (\dfrac{7}{4} \right ) \qquad &&\text{Solve for x} \end{align*}\]

Exercise \(\PageIndex{7}\)

Solve \(3+e^{2t}=7e^{2t}\).

\(t=\ln \left (\dfrac{1}{\sqrt{2}} \right )=−\dfrac{1}{2}\ln(2)\)

Extraneous Solutions

Sometimes the methods used to solve an equation introduce an extraneous solution , which is a solution that is correct algebraically but does not satisfy the conditions of the original equation. One such situation arises in solving when the logarithm is taken on both sides of the equation. In such cases, remember that the argument of the logarithm must be positive. If the number we are evaluating in a logarithm function is negative, there is no output.

Example \(\PageIndex{8}\): Solving Exponential Functions in Quadratic Form

Solve \(e^{2x}−e^x=56\).

\[\begin{align*} e^{2x}-e^x&= 56\\ e^{2x}-e^x-56&= 0 \qquad &&\text{Get one side of the equation equal to zero}\\ (e^x+7)(e^x-8)&= 0 \qquad &&\text{Factor by the FOIL method}\\ e^x+7&= 0 \qquad &&\text{or} \\ e^x-8&= 0 \qquad &&\text{If a product is zero, then one factor must be zero}\\ e^x&= -7 \qquad &&\text{or} \\ e^x&= 8 \qquad &&\text{Isolate the exponentials}\\ e^x&= 8 \qquad &&\text{Reject the equation in which the power equals a negative number}\\ x&= \ln8 \qquad &&\text{Solve the equation in which the power equals a positive number} \end{align*}\]

When we plan to use factoring to solve a problem, we always get zero on one side of the equation, because zero has the unique property that when a product is zero, one or both of the factors must be zero. We reject the equation \(e^x=−7\) because a positive number never equals a negative number. The solution \(\ln(−7)\) is not a real number, and in the real number system this solution is rejected as an extraneous solution.

Exercise \(\PageIndex{8}\)

Solve \(e^{2x}=e^x+2\).

Q&A: Does every logarithmic equation have a solution?

No. Keep in mind that we can only apply the logarithm to a positive number. Always check for extraneous solutions.

Using the Definition of a Logarithm to Solve Logarithmic Equations

We have already seen that every logarithmic equation \({\log}_b(x)=y\) is equivalent to the exponential equation \(b^y=x\). We can use this fact, along with the rules of logarithms, to solve logarithmic equations where the argument is an algebraic expression.

For example, consider the equation \({\log}_2(2)+{\log}_2(3x−5)=3\). To solve this equation, we can use rules of logarithms to rewrite the left side in compact form and then apply the definition of logs to solve for \(x\):

\[\begin{align*} {\log}_2(2)+{\log}_2(3x-5)&= 3\\ {\log}_2(2(3x-5))&= 3 \qquad \text{Apply the product rule of logarithms}\\ {\log}_2(6x-10)&= 3 \qquad \text{Distribute}\\ 2^3&= 6x-10 \qquad \text{Apply the definition of a logarithm}\\ 8&= 6x-10 \qquad \text{Calculate } 2^3\\ 18&= 6x \qquad \text{Add 10 to both sides}\\ x&= 3 \qquad \text{Divide by 6} \end{align*}\]

USING THE DEFINITION OF A LOGARITHM TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expression \(S\) and real numbers \(b\) and \(c\),where \(b>0\), \(b≠1\),

\[\begin{align} {\log}_b(S)=c \text{ if and only if } b^c=S \end{align}\]

Example \(\PageIndex{9}\): Using Algebra to Solve a Logarithmic Equation

Solve \(2\ln x+3=7\).

\[\begin{align*} 2\ln x+3&= 7\\ 2\ln x&= 4 \qquad \text{Subtract 3}\\ \ln x&= 2 \qquad \text{Divide by 2}\\ x&= e^2 \qquad \text{Rewrite in exponential form} \end{align*}\]

Exercise \(\PageIndex{9}\)

Solve \(6+\ln x=10\).

Example \(\PageIndex{10}\): Using Algebra Before and After Using the Definition of the Natural Logarithm

Solve \(2\ln(6x)=7\).

\[\begin{align*} 2\ln(6x)&= 7\\ \ln(6x)&= \dfrac{7}{2} \qquad \text{Divide by 2}\\ 6x&= e^{\left (\dfrac{7}{2} \right )} \qquad \text{Use the definition of }\ln \\ x&= \dfrac{1}{6}e^{\left (\dfrac{7}{2} \right )} \qquad \text{Divide by 6} \end{align*}\]

Exercise \(\PageIndex{10}\)

Solve \(2\ln(x+1)=10\).

\(x=e^5−1\)

Example \(\PageIndex{11}\): Using a Graph to Understand the Solution to a Logarithmic Equation

Solve \(\ln x=3\).

\[\begin{align*} \ln x&= 3\\ x&= e^3 \qquad \text{Use the definition of the natural logarithm} \end{align*}\]

Figure \(\PageIndex{3}\) represents the graph of the equation. On the graph, the x -coordinate of the point at which the two graphs intersect is close to \(20\). In other words \(e^3≈20\). A calculator gives a better approximation: \(e^3≈20.0855\).

Graph of two questions, y=3 and y=ln(x), which intersect at the point (e^3, 3) which is approximately (20.0855, 3).

Exercise \(\PageIndex{11}\)

Use a graphing calculator to estimate the approximate solution to the logarithmic equation \(2^x=1000\) to \(2\) decimal places.

\(x≈9.97\)

Using the One-to-One Property of Logarithms to Solve Logarithmic Equations

As with exponential equations, we can use the one-to-one property to solve logarithmic equations. The one-to-one property of logarithmic functions tells us that, for any real numbers \(x>0\), \(S>0\), \(T>0\) and any positive real number \(b\), where \(b≠1\),

\({\log}_bS={\log}_bT\) if and only if \(S=T\).

For example,

If \({\log}_2(x−1)={\log}_2(8)\), then \(x−1=8\).

So, if \(x−1=8\), then we can solve for \(x\),and we get \(x=9\). To check, we can substitute \(x=9\) into the original equation: \({\log}_2(9−1)={\log}_2(8)=3\). In other words, when a logarithmic equation has the same base on each side, the arguments must be equal. This also applies when the arguments are algebraic expressions. Therefore, when given an equation with logs of the same base on each side, we can use rules of logarithms to rewrite each side as a single logarithm. Then we use the fact that logarithmic functions are one-to-one to set the arguments equal to one another and solve for the unknown.

For example, consider the equation \(\log(3x−2)−\log(2)=\log(x+4)\). To solve this equation, we can use the rules of logarithms to rewrite the left side as a single logarithm, and then apply the one-to-one property to solve for \(x\):

\[\begin{align*} \log(3x-2)-\log(2)&= \log(x+4)\\ \log \left (\dfrac{3x-2}{2} \right )&= \log(x+4) \qquad \text{Apply the quotient rule of logarithms}\\ \dfrac{3x-2}{2}&= x+4 \qquad \text{Apply the one to one property of a logarithm}\\ 3x-2&= 2x+8 \qquad \text{Multiply both sides of the equation by 2}\\ x&= 10 \qquad \text{Subtract 2x and add 2} \end{align*}\]

To check the result, substitute \(x=10\) into \(\log(3x−2)−\log(2)=\log(x+4)\).

\[\begin{align*} \log(3(10)-2)-\log(2)&= \log((10)+4) \\ \log(28)-\log(2)&= \log(14)\\ \log \left (\dfrac{28}{2} \right )&= \log(14) \qquad \text{The solution checks} \end{align*}\]

USING THE ONE-TO-ONE PROPERTY OF LOGARITHMS TO SOLVE LOGARITHMIC EQUATIONS

For any algebraic expressions \(S\) and \(T\) and any positive real number \(b\), where \(b≠1\),

Note, when solving an equation involving logarithms, always check to see if the answer is correct or if it is an extraneous solution.

How to: Given an equation containing logarithms, solve it using the one-to-one property

  • Use the rules of logarithms to combine like terms, if necessary, so that the resulting equation has the form \({\log}_bS={\log}_bT\).
  • Use the one-to-one property to set the arguments equal.

Example \(\PageIndex{12}\): Solving an Equation Using the One-to-One Property of Logarithms

Solve \(\ln(x^2)=\ln(2x+3)\).

\[\begin{align*} \ln(x^2)&= \ln(2x+3)\\ x^2&= 2x+3 \qquad \text{Use the one-to-one property of the logarithm}\\ x^2-2x-3&= 0 \qquad \text{Get zero on one side before factoring}\\ (x-3)(x+1)&= 0 \qquad \text{Factor using FOIL}\\ x-3&= 0 \qquad \text{or } x+1=0 \text{ If a product is zero, one of the factors must be zero}\\ x=3 \qquad \text{or} \\ x&= -11 \qquad \text{Solve for x} \end{align*}\]

There are two solutions: \(3\) or \(−1\). The solution \(−1\) is negative, but it checks when substituted into the original equation because the argument of the logarithm functions is still positive.

Exercise \(\PageIndex{12}\)

Solve \(\ln(x^2)=\ln1\).

\(x=1\) or \(x=−1\)

Solving Applied Problems Using Exponential and Logarithmic Equations

In previous sections, we learned the properties and rules for both exponential and logarithmic functions. We have seen that any exponential function can be written as a logarithmic function and vice versa. We have used exponents to solve logarithmic equations and logarithms to solve exponential equations. We are now ready to combine our skills to solve equations that model real-world situations, whether the unknown is in an exponent or in the argument of a logarithm.

One such application is in science, in calculating the time it takes for half of the unstable material in a sample of a radioactive substance to decay, called its half-life . Table \(\PageIndex{1}\) lists the half-life for several of the more common radioactive substances.

We can see how widely the half-lives for these substances vary. Knowing the half-life of a substance allows us to calculate the amount remaining after a specified time. We can use the formula for radioactive decay:

\[\begin{align} A(t)&= A_0e^{\tfrac{\ln(0.5)}{T}t}\\ A(t)&= A_0e^{\tfrac{\ln(0.5)t}{T}}\\ A(t)&= A_0{(e^{\ln(0.5)})}^{\tfrac{t}{T}}\\ A(t)&= A_0{\left (\dfrac{1}{2}\right )}^{\tfrac{t}{T}}\\ \end{align}\]

  • \(A_0\) is the amount initially present
  • \(T\) is the half-life of the substance
  • \(t\) is the time period over which the substance is studied
  • \(y\) is the amount of the substance present after time \(t\)

Example \(\PageIndex{13}\): Using the Formula for Radioactive Decay to Find the Quantity of a Substance

How long will it take for ten percent of a \(1000\)-gram sample of uranium-235 to decay?

\[\begin{align*} y&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t}\\ 900&= 1000e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{After } 10\% \text{ decays, 900 grams are left}\\ 0.9&= e^{\tfrac{\ln(0.5)}{703,800,000}t} \qquad \text{Divide by 1000}\\ \ln(0.9)&= \ln \left (e^{\tfrac{\ln(0.5)}{703,800,000}t} \right ) \qquad \text{Take ln of both sides}\\ \ln(0.9)&= \dfrac{\ln(0.5)}{703,800,000}t \qquad \ln(e^M)=M\\ t&= 703,800,000\times \dfrac{\ln(0.9)}{\ln(0.5)} \qquad \text{years Solve for t}\\ t&\approx 106,979,777 \qquad \text{years} \end {align*} \]

Ten percent of \(1000\) grams is \(100\) grams. If \(100\) grams decay, the amount of uranium-235 remaining is \(900\) grams.

Exercise \(\PageIndex{13}\)

How long will it take before twenty percent of our 1000-gram sample of uranium-235 has decayed?

\(t=703,800,000×\dfrac{\ln(0.8)}{\ln(0.5)}\)years ≈ 226,572,993 years.

Access these online resources for additional instruction and practice with exponential and logarithmic equations.

  • Solving Logarithmic Equations
  • Solving Exponential Equations with Logarithms

Key Equations

Key concepts.

  • We can solve many exponential equations by using the rules of exponents to rewrite each side as a power with the same base. Then we use the fact that exponential functions are one-to-one to set the exponents equal to one another and solve for the unknown.
  • When we are given an exponential equation where the bases are explicitly shown as being equal, set the exponents equal to one another and solve for the unknown. See Example \(\PageIndex{1}\).
  • When we are given an exponential equation where the bases are not explicitly shown as being equal, rewrite each side of the equation as powers of the same base, then set the exponents equal to one another and solve for the unknown. See Example \(\PageIndex{2}\), Example \(\PageIndex{3}\), and Example \(\PageIndex{4}\).
  • When an exponential equation cannot be rewritten with a common base, solve by taking the logarithm of each side. See Example \(\PageIndex{5}\).
  • We can solve exponential equations with base \(e\),by applying the natural logarithm of both sides because exponential and logarithmic functions are inverses of each other. See Example \(\PageIndex{6}\) and Example \(\PageIndex{7}\).
  • After solving an exponential equation, check each solution in the original equation to find and eliminate any extraneous solutions. See Example \(\PageIndex{8}\).
  • When given an equation of the form \({\log}_b(S)=c\), where \(S\) is an algebraic expression, we can use the definition of a logarithm to rewrite the equation as the equivalent exponential equation \(b^c=S\), and solve for the unknown. See Example \(\PageIndex{9}\) and Example \(\PageIndex{10}\).
  • We can also use graphing to solve equations with the form \({\log}_b(S)=c\). We graph both equations \(y={\log}_b(S)\) and \(y=c\) on the same coordinate plane and identify the solution as the x- value of the intersecting point. See Example \(\PageIndex{11}\).
  • When given an equation of the form \({\log}_bS={\log}_bT\), where \(S\) and \(T\) are algebraic expressions, we can use the one-to-one property of logarithms to solve the equation \(S=T\) for the unknown. See Example \(\PageIndex{12}\).
  • Combining the skills learned in this and previous sections, we can solve equations that model real world situations, whether the unknown is in an exponent or in the argument of a logarithm. See Example \(\PageIndex{13}\).

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If you're taking a high school or college math class, you'll likely cover natural logs. But what are natural logs? What is ln? Why does the letter e keep showing up?

Natural logs may seem difficult, but once you understand a few key natural log rules, you'll be able to easily solve even very complicated-looking problems. In this guide, we explain the four most important natural logarithm rules, discuss other natural log properties you should know, go over several examples of varying difficulty, and explain how natural logs differ from other logarithms.

What Is ln?

The natural log, or ln, is the inverse of e . The letter ‘ e' represents a mathematical constant also known as the natural exponent. Like π, e is a mathematical constant and has a set value. The value of e is equal to approximately 2.71828.

e appears in many instances in mathematics, including scenarios about compound interest, growth equations, and decay equations. ln( x ) is the time needed to grow to x , while e x is the amount of growth that has occurred after time x .

Because e is used so commonly in math and economics, and people in these fields often need to take the logarithm with a base of e of a number to solve an equation or find a value, the natural log was created as a shortcut way to write and calculate log base e . The natural log simply lets people reading the problem know that you're taking the logarithm, with a base of e , of a number. So ln( x ) = log e ( x ). As an example, ln( 5 ) = log e ( 5 ) = 1.609.

The 4 Key Natural Log Rules

There are four main rules you need to know when working with natural logs, and you'll see each of them again and again in your math problems. Know these well because they can be confusing the first time you see them, and you want to make sure you have basic rules like these down solid before moving on to more difficult logarithm topics.

Product Rule

  • ln(x)(y) = ln(x) + ln(y)
  • The natural log of the multiplication of x and y is the sum of the ln of x and ln of y.
  • Example: ln(8)(6) = ln(8) + ln(6)

Quotient Rule

  • ln(x/y) = ln(x) - ln(y)
  • The natural log of the division of x and y is the difference of the ln of x and ln of y.
  • Example: ln(7/4) = ln(7) - ln(4)

Reciprocal Rule

  • ln(1/x) = −ln(x)
  • The natural log of the reciprocal of x is the opposite of the ln of x.
  • Example: ln(⅓)= -ln(3)
  • ln( x y ) = y * ln(x)
  • The natural log of x raised to the power of y is y times the ln of x.
  • Example: ln(5 2 ) = 2 * ln(5)

body_logarithm

Key Natural Log Properties

In addition to the four natural logarithm rules discussed above, there are also several ln properties you need to know if you're studying natural logs. Have these memorized so you can quickly move onto the next step of the problem without wasting time trying to remember common ln properties.

As you can see from the final three rows, ln( e )=1, and this is true even if one is raised to the power of the other. This is because the ln and e are inverse functions of each other.

Natural Log Sample Problems

Now it's time to put your skills to the test and ensure you understand the ln rules by applying them to example problems. Below are three sample problems. Try to work them out on your own before reading through the explanation.

Evaluate ln(7 2 /5)

First, we use the quotient rule to get: ln(7 2 ) - ln(5).

Next, we use the power rule to get: 2ln(7) -ln(5).

If you don't have a calculator, you can leave the equation like this, or you can calculate the natural log values: 2(1.946) - 1.609 = 3.891 - 1.609 = 2.282.

Evaluate ln( e ) /7

For this problem, we need to remember than ln( e )=1

This means the problem simplifies to 1/7, which is our answer

Solve ln (5 x -6)=2

When you have multiple variables within the ln parentheses, you want to make e the base and everything else the exponent of e . Then you'll get ln and e next to each other and, as we know from the natural log rules, e ln(x) =x.

So, the equation becomes e ln(5x-6) = e 2

Since e ln(x) = x , e ln(5x-6) = 5x-6

Therefore 5 x -6= e 2

Since e is a constant, you can then figure out the value of e 2 , either by using the e key on your calculator or using e's estimated value of 2.718.

5 x -6 =7.389

Now we'd add 6 to both sides

5 x = 13.389

Finally, we'd divide both sides by 5.

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How Are Natural Logs Different From Other Logarithms?

As a reminder, a logarithm is the opposite of a power. If you take the log of a number, you're undoing the exponent. The key difference between natural logs and other logarithms is the base being used. Logarithms typically use a base of 10 (although it can be a different value, which will be specified), while natural logs will always use a base of e .

This means ln(x)=log e ( x )

If you need to convert between logarithms and natural logs, use the following two equations:

  • log 10 ( x ) = ln(x) / ln(10)
  • ln(x) = log 10 ( x ) / log 10 ( e )

Other than the difference in the base (which is a big difference) the logarithm rules and the natural logarithm rules are the same:

Summary: Natural Log Rules

The natural log, or ln, is the inverse of e. The rules of natural logs may seem counterintuitive at first, but once you learn them they're quite simple to remember and apply to practice problems.

The four main ln rules are:

  • ln(x)( y) = ln(x) + ln(y)
  • ln(1/x)=−ln(x)
  • n( x y ) = y*ln(x)

The key difference between natural logs and other logarithms is the base being used.

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Christine graduated from Michigan State University with degrees in Environmental Biology and Geography and received her Master's from Duke University. In high school she scored in the 99th percentile on the SAT and was named a National Merit Finalist. She has taught English and biology in several countries.

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The Ultimate Guide to Solving Logarithmic Equations in Algebra

Equations containing variables in logarithmic expressions are called logarithmic equations (sometimes shortened as “log equations”). Solving logarithmic equations can be easy and entertaining if you are aware of the principal methods and different scenarios. Here, we’ll provide a comprehensive guide on the most efficient methods to solve log equations.

Type 1 Logarithmic Equations

The simplest logarithmic equations are equations of the form

where the base of the logarithm, b, is a positive number, b ≠ 1. For any real value of the variable x, this equation has a single solution:

For example, the log equation

only has one solution:

Consider logarithmic equations of the form

This can be simplified by using the change of base formula for logarithms:

This results in

can be written in the following form:

This has the solution

Type 2 Logarithmic Equations

Consider a slightly more complicated logarithmic equation:

Here, again, the base of the logarithm, b, is a positive number, b ≠ 1, and f(x) is some elementary algebraic function. This equation can be solved by introducing a new variable, t = f(x), defined as

Thus, we can write the following general solution for this equation:

Let’s illustrate this idea with examples.

Say we want to solve the following logarithmic equation:

We substitute the following variable:

This turns the equation into a standard form:

Thus, we have a quadratic equation for the variable x:

We can easily determine the roots of this quadratic equation, which are also the solutions of the original logarithmic equation:

Consider the following log equation:

This can be solved after applying the change of base formula:

We must also substitute a new variable:

For the variable t, we have

This results in t = 10. Thus, we obtain

Both roots of this algebraic equation are solutions to the original logarithmic equation:

This method also works if f(x) is a logarithmic function by itself. For example, consider the equation

As usual, we can introduce a new variable t, according to the following equation:

Then, t is a solution to the simple log equation

which can be solved easily:

The variable x also becomes the solution to a simple log equation:

Finally, we can find the solution:

Type 3 Logarithmic Equations

Next, we will investigate how to solve log equations of the form

where f(x) and h(x) are some elementary algebraic functions, and b is a positive number, b ≠ 1. This log equation is equivalent to the algebraic equation

We should also remember that the domain of any logarithmic function is nonnegative, real numbers. Thus, among all the solutions to the equation f(x) = h(x), we should only select solutions that satisfy one of the following conditions:

This guarantees that the logarithmic functions are well-defined.

As an example, let us consider the equation

Among all the roots of this equation, which can be defined as

Only those that satisfy the condition

can be solutions of the original logarithmic equation. Hence, roots x 1 and x 2 should be rejected, and the only solution is

If we take into account that

We can rewrite this log equation as

Applying the method described for this type of logarithmic equation, we obtain

This equation is equivalent to

This has a single solution:

Obviously, this solution satisfies the condition x > 0.

The method we are now investigating only requires a slight modification if the logarithmic expressions on both sides of the log equation have different bases:

where both a and b are positive numbers, a ≠ 1 and b ≠1. We can rewrite the left-hand side of this equation in the form

Thus, we have

Then, the log equation in question is again equivalent to the following algebraic equation:

Pay attention to the fact that both of the following conditions must be satisfied in this case:

To illustrate, consider the log equation

Using the change of base formula, we have

Thus, we can write

This results in the following algebraic equation:

Further, remember that both initial logarithmic functions are defined only in the region

Thus, between the two roots of the quadratic equation

only x = 2 is a solution to the original log equation.  

Type 4 Logarithmic Equations

A more challenging class of logarithmic equations are equations of the form

where b is a positive number, b ≠ 1, and

These are some algebraic functions (some of them can be constant numbers).

Solving logarithmic equations of this type is equivalent to solving the following system of algebraic equations:

For example, consider the following log equation:

As we have just discussed, to solve this logarithmic equation, we have to solve

Among the two roots of the quadratic equation,

only the second one satisfies the above inequalities. Thus, the only solution of the given logarithmic equation is

Based on our general consideration that solving this logarithmic equation is equivalent to solving a mixed system of algebraic equations,

These, in turn, are equivalent to the system

Both roots of the quadratic equation,

satisfy the inequality x > 1/3. Thus, both roots are solutions of the logarithmic equation.

Consider the following equation:

This belongs to the same class of logarithmic equations. It is not difficult to observe that it can be written in the form

This equation is equivalent to a mixed system of algebraic equations:

The quadratic equation can be written in the form

This has two roots:

We can easily verify that the first root, x 1 = – 1, does not satisfy the inequality 2 x + 1 > 0, while the second root, x 2 = 1/2, satisfies both inequalities. Thus,

This is the only solution to the logarithmic equation in question.

If we encounter a problem with fractions of logarithms, most likely, the log equation can be transformed into the same standard form. For example, if we need to solve

First, we note that

This is the range for which the variable x is valid. Next, rewrite the equation in the form

Thus, the logarithmic equation under investigation becomes an algebraic equation:

This can be solved easily:

Finally, the only root satisfying the condition x > 0 is

Type 5 Logarithmic Equations

Finally, we will learn how to solve logarithmic equations of the form

Here, h(x) is some logarithmic function and F(u) is an elementary algebraic function. In this case, we can introduce a new variable t = h(x) and solve:

be the n real numbers that are solutions to the algebraic equation F(t) = 0. Then, to solve the original logarithmic equation, we have to find solutions to the following system of n algebraic equations:

Using the fact that

we can write this log equation in the following form:

For the auxiliary variable

we obtain a simple algebraic equation:

This has the following roots:

Thus, we have to solve two logarithmic equations:

This is quite an easy task. The corresponding solutions are

These are both in the range of validity for the logarithmic function, x > 0.

Consider another logarithmic equation:

we can rewrite this equation in the form

Now, we introduce a new variable:

This is the solution of the equation

After some simple algebraic transformations, we obtain

Thus, we have to solve the following log equations:

The obvious solutions are

Although this log equation seems quite different from those we have considered so far, we will now show how it can be solved using the same methods.

First, since the base of a logarithm can only be a positive number not equal to unity, the domain of validity for the variable x can be expressed as

Now, we can use a change of base formula for logarithms to express log x 3 in terms of log 3 x:

Analogously, we can write

The original logarithmic equation now transforms into an equation of standard form:

This can be solved by the following substitution:

The equation in terms of the variable t takes the form:

We can easily find the corresponding roots:

Thus, we have two log equations:

These are extremely easy to solve, yielding the following results:

Both solutions are in the valid range for the variable x.

Hopefully, this review article has improved your understanding of how to solve logarithmic equations in algebra. If you work through the wide range of examples presented here, you will be prepared to solve logarithmic equations of any difficulty. Best of luck!

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how to solve complicated logs

Inside the race to grasp the fate of the Colorado River

For decades, officials have tried to predict the river’s future flows. now they’re hoping innovative, web-based tools will help salvage the lifeline it provides..

how to solve complicated logs

BOULDER, Colo. — To ensure that the Colorado River can remain a lifeline for 40 million people, the federal government is looking for answers in the extremes of the distant past and the warnings of a hotter future.

In a low-slung building at the University of Colorado at Boulder, a group of engineers and scientists have developed a cutting-edge approach to help negotiators fashion the next major deal to divvy up the dwindling river for decades to come.

Despite significant rainfall in recent months, Lake Mead could return to near-historic lows by 2025. As the Bureau of Reclamation looks to reach a deal by the end of the year — before a potential change in administration — the agency is, for the first time, putting climate change at the center of how it’s planning the future.

Those who rely on the river are now testing water-sharing strategies with the agency’s new web-based tool that harnesses more than 8,000 possible futures of the river to see how policies stand up against the wild swings and uncertainties brought on by the warming climate.

“We ultimately get a very wide range of conditions that could happen under climate change,” Rebecca Smith, a Reclamation official, said during a November seminar. “And scientists don’t expect that to be narrowing anytime soon.”

The force driving this innovation is a chilling one: Policymakers’ recognition that the way they’ve forecast the river’s future no longer works.

By relying on records of the river’s flow over the last century, the federal government and western states repeatedly underestimated the drought and failed to keep major reservoirs from nearing dangerous levels that could threaten the water supply for millions of people. Over the years, Reclamation has looked to climate science to solve the real world problem of a shrinking Colorado River, according to river experts and people involved in the effort. Even so, the deals struck with states over the past two decades to cope with the drought did not incorporate climate change models into the simulations of the river used to set long-term policy.

This is changing now. Reclamation’s new approach will test policies against a future informed by climate models that project warming to continue, as well as tree ring records of ancient droughts far worse than anything in the recent past — incorporating a much wider range of possible river flows than available in the recent historical record.

“We’re teetering right on the edge here of being able to say, yep, climate change models are a part of our decision-making,” said Terry Fulp, who spent three decades with Reclamation, including eight years as the Lower Colorado River Basin regional director until retiring in 2020. “It’s been a long time coming.”

To do this, a wonky band of government officials and academics are diving into a new world of unpredictability.

Their approach is called “decision-making under deep uncertainty” and it throws out the notion that anyone can predict what the future flow of the river might be — a strategy that led to Reclamation being in a “crisis management mode,” said one official not authorized to speak publicly.

The most recent crisis eased about a year ago. The nation’s second-largest reservoir, Lake Powell, nearly dropped to the point where its hydroelectric dam could no longer produce power. It was saved by an unusually wet winter — and a short-term deal with states to conserve water in exchange for billions in federal money. But scientists warn that renewed drought could quickly threaten the region again.

“Last year was maybe the anomaly,” said Tom Buschatzke, director of the Arizona Department of Water Resources.

Rethinking past assumptions

When Fulp joined Reclamation in 1989, the future of the Colorado River was planned on a single computer in a suburban Denver office building.

The lumbering mainframe at the Lakewood office took up an entire room and slowly churned out rows of numbers on perforated paper — a simulation of how the most important river in the West would be distributed among its dozen major reservoirs for years into the future.

The only people who could use this model — known as the Colorado River Simulation System, or CRSS, were federal government water managers such as Fulp. And to make tweaks required rewriting its code in the now-archaic computer language called Fortran. It was, Fulp recalled, “really cumbersome.”

In the early 1990s, Fulp teamed up with academics at UC Boulder’s Center for Advanced Decision Support for Water and Environmental Systems to develop software known as RiverWare and a version of the CRSS model that could be used by others reliant on the river, from states and cities to farmers and tribes.

In the Colorado River basin, the states were competing more for water; new federal environmental laws required comparing alternatives before making resource decisions; and many distrusted the federal government.

“They all hated Reclamation,” recalled Edith Zagona, a former Reclamation official who directs the UC Boulder Center. “We had to prove to them that this newfangled software could work. They didn’t want anyone pulling a fast one on them.”

Over the years, the software and the models running on it became a fundamental tool for settling disputes, avoiding litigation among rival parties and reaching agreements on how the water could be stored and distributed. It’s used on several major rivers in the United States, including the Columbia River and the Rio Grande.

The CRSS model simulates the flow of water through canals, pipelines and reservoirs as the Colorado traverses 1,450 miles, seven states and two countries — from the snow-capped peaks of the Rocky Mountains to the Sea of Cortez. Programmed into it are the rules that govern who gets priority access to that water — the complex web of regulations, statutes, treaties and court cases that date back more than a century, known as the Law of the River.

It can reveal how changes in water supply, demand or the policy environment will impact issues including reservoir levels, hydropower production, flood control or protecting the endangered fish that swim through the Grand Canyon.

To run it, water managers must estimate the streamflow entering the river system. Since the model was first developed, they’ve based this on what had been observed on the Colorado River dating back to 1906.

Scientists knew this period included very wet years in the 1920s, when the western water sharing deal was forged, and in the 1980s, when Lake Powell almost overflowed the Glen Canyon Dam. But the 20th century record was seen as the best way to understand what the river was capable of doing.

“It was almost universal in water planning: the assumption that the future would look like the past,” said Jeff Lukas, a climatologist and Colorado River researcher.

But after the drought started in 1999 — and Lake Mead and Lake Powell commenced their long decline — there were growing doubts about the usefulness of the 20th century record. One major warning came in the form of tree rings.

The rings inside logs, stumps and standing dead trees told a much older story of how much water was available in the Colorado River basin. Researchers in 2007 created a record that dated back to 762 A.D. that revealed punishing Medieval droughts, including a 62-year dry stretch in the 1100s that produced streamflows lower than anything in the 20th century.

With levels at Lake Mead and Lake Powell dropping from nearly full, in 2000, to about half full five years later, Reclamation and the states negotiated an agreement over how to operate the reservoirs as they declined — and how shortages would be divvied up.

In 2007, the modeling predicted a zero percent chance that Lake Powell’s elevation would fall to the point that it couldn’t help produce power by 2026. Last year, states struck an emergency deal to make unprecedented cuts in water use to avoid that very fate.

Charting the river’s future

In 2012, after a decade of drought, Reclamation published a major study assessing the health of the Colorado River and what the warming climate would mean for its future.

It wasn’t a simple answer. The leading global climate models needed to be adapted to the Colorado River region — where topography in the Rocky Mountains varied widely — then combined with other models to estimate stream flows. While the results consistently projected higher temperatures for the region, with less snowpack in the Rockies and more frequent droughts, it was — and remains — less clear how much rain and snow would fall each year.

“The precipitation part of climate model output is not good,” said Brad Udall, a water and climate scientist at Colorado State University. “It still continues to have problems, especially on a regional basis, getting things right.”

A Rand Corp. scientist named David Groves, who had been hired to help on the 2012 study, saw this uncertainty as an opportunity to rethink how river modeling was done. Instead of trying to make possibly incorrect predictions about what the Colorado River would look like in the future, he and his colleagues advocated to test different river management strategies against thousands of possible futures of the river derived from a wide range of sources — the tree ring record, climate models, the historical record — to see how well they held up.

At the time, such intensive computer modeling was slow. In November 2014, Groves convened a group of Colorado River stakeholders at the Lawrence Livermore National Laboratory in California to show how the lab’s supercomputers could accelerate the process. He tested policies against 12,000 river scenarios — something that would normally take six weeks.

“While we fed them lunch, we ran the supercomputer and it did the whole analysis in an hour,” Groves said.

Back then, he said, there were plenty of ominous possible outcomes in his analysis, including that Lake Mead would disappear if consumption patterns continued.

“People saw it and it worried them,” Groves recalled. “But even if someone says this is what could happen, you can still say, ‘Well, that’s far off.’”

By the fall of 2022, Lake Mead and Lake Powell were just a quarter full. And Reclamation was predicting a more than 50 percent chance these lakes would drop below critical thresholds without further action, although forecasts are now less dire.

Trying to avoid disaster

The new web tool, built by Virga Labs, a water consulting firm, in partnership with Zagona’s center and Reclamation, is the culmination of years of work on the science of “decision-making under deep uncertainty.” It uses cloud-computing to run the CRSS model in thousands of variations in ways that haven’t been possible in the past, said Virga Labs chief executive Season Martin.

But the tool also represents a more fundamental change in how Reclamation looks at the problem of a dwindling Colorado River. Instead of trying to assign probabilities about future risk, the agency aims to test different ways to manage the river against the widest possible range of river scenarios — and see what works best to avoid disaster.

“We want to know that things are going to be okay even in really challenging conditions,” Smith, the Reclamation official, said in November.

Buschatzke, Arizona’s top water official, called the new modeling tool a “very positive step forward,” because it allows people or entities without the technical expertise to explore changes in river management. But he expects his agency to continue to run numbers and test strategies the old way, as well.

“We’ve spent enough time and effort learning how to do it,” he said.

Reclamation said it will also use other analytical methods it has relied on in the past as it assesses the proposals that states are expected to present in March on how to share a river stressed by climate change.

Not everyone is confident this new approach will deliver better results — or keep reservoirs from running dry in the future.

Udall, the Colorado State University scientist, said sophisticated new tools may give the illusion of being able to find a safe path forward for a river whose natural flow is down 20 percent in the past two decades. Simply cutting the amount of water states can take by a big amount may be wiser, he said.

“I think we get too clever by half with all this technology,” he said. “I think you can make this process a whole lot simpler. Which is we need to plan for a lot lower flows, and let’s not hide behind or obscure some of these tough details with really complicated models.”

how to solve complicated logs

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  3. how to solve complicated logs

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    how to solve complicated logs

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    how to solve complicated logs

  6. How to solve complicated logarithmic equations?

    how to solve complicated logs

VIDEO

  1. C1 Common Log, Natural Log and other Logarithmic Rules

  2. LOGARITHM TABLES. EVALUATION OF NUMERICAL PROBLEMS USING LOGS

  3. Log Equation with Different Bases

  4. Logarithm rules

  5. How to solve complicated quadratic equation

  6. Solve This Logarithm Problem Step-by-Step

COMMENTS

  1. Solving Complex Logarithmic Equations

    This video explains how to solve complex logarithmic equations using properties of logarithms such as the change of base formula, the power rule, and other s...

  2. Algebra

    Show All Solutions Hide All Solutions a 2log9(√x)−log9(6x −1) = 0 2 log 9 ( x) − log 9 ( 6 x − 1) = 0 Show Solution b logx +log(x−1) =log(3x+12) log x + log ( x − 1) = log ( 3 x + 12) Show Solution c ln10−ln(7 −x) = lnx ln 10 − ln ( 7 − x) = ln x Show Solution

  3. Evaluate logarithms (advanced) (practice)

    Course: Algebra 2 > Unit 8. Lesson 1: Introduction to logarithms. Intro to logarithms. Intro to Logarithms. Evaluate logarithms. Evaluating logarithms (advanced) Evaluate logarithms (advanced) Relationship between exponentials & logarithms. Relationship between exponentials & logarithms: graphs.

  4. Solving More Complex Logarithmic Equations

    Solving More Complex Logarithmic Equations Sometimes logarithmic equations are more complex. Let's examine a few of these cases: 1. If you have the same logarithm on both sides, their arguments will equal each other. Example: Since the base of the natural log is e, we will raise both sides to be powers of e.

  5. Solving Logarithmic Equations

    Example 1: Solve the logarithmic equation. Since we want to transform the left side into a single logarithmic equation, we should use the Product Rule in reverse to condense it. Here is the rule, just in case you forgot. Given Apply Product Rule from Log Rules. Distribute: [latex]\left ( {x + 2} \right)\left ( 3 \right) = 3x + 6 [/latex]

  6. Solving (Challenging) Log Equations Different Bases

    © 2023 Google LLC Learn how to Solve (Challenging) Log Equations with Different Bases in this free math video tutorial by Mario's Math Tutoring. In this video we look at a mor...

  7. Solving a complicated logarithmic equation

    In this video, I showed how to solve a complicated logarithmic equation using basic rules of logarithm

  8. 3 Ways to Solve Logarithms

    1 Know the logarithm definition. Before you can solve logarithms, you need to understand that a logarithm is essentially another way to write an exponential equation. It's precise definition is as follows: y = logb (x) If and only if: by = x Note that b is the base of the logarithm. It must also be true that: b > 0 b does not equal 1

  9. Solving Logarithmic Equations

    For many equations with logarithms, solving them is simply a matter of using the definition of \log x logx to eliminate logarithms from the equation and convert it into a polynomial or exponential equation. Find x x if \log_2 (3x+1) = 4 log2(3x+1) = 4. By the definition of the logarithm,

  10. Logarithms

    They allow us to solve challenging exponential equations, and they are a good excuse to dive deeper into the relationship between a function and its inverse. ... Solve exponential equations using logarithms: base-2 and other bases Get 3 of 4 questions to level up! Solving exponential models. Learn. Exponential model word problem: medication ...

  11. How to solve complicated logarithmic equations?

    Explanation: These questions are tricky because you have a constant in between all of those logs. To get around this, turn them into logs so in this case: 1 = (log10)10 Once you've done this, youcan use your other log laws to solve the equations: 2log5 becomes log25 and when you add logs, you multiply the brackets: log25 +log(x +1) = log(25(x + 1))

  12. Logarithmic Equations: Very Difficult Problems with Solutions

    Difficult Logarithmic Equations: Very Difficult Problems with Solutions Problem 1 Find the root of the equation \displaystyle 2+lg\sqrt {1+x}+3lg\sqrt {1-x}=lg\sqrt {1-x^2} 2+lg 1 +x +3lg 1−x = lg 1−x2 \displaystyle \frac {9} {100} 1009 \displaystyle \frac {99} {100} 10099 \displaystyle \frac {9} {10} 109 \displaystyle \frac {1} {9} 91

  13. Logarithmic equations: variable in the argument

    11 years ago The human ear works as a logarithmic function. The tempered musical scale is exponential so after passing through a logarithmic function (ear) it become linear. This mix of functions makes the transition from notes of the scale perceived by our brain softly as if the notes were located exactly one after the other.

  14. Logarithms Calculator

    Free Logarithms Calculator - Simplify logarithmic expressions using algebraic rules step-by-step

  15. 4.7: Exponential and Logarithmic Equations

    Use the rules of exponents to simplify, if necessary, so that the resulting equation has the form bS = bT. Use the one-to-one property to set the exponents equal. Solve the resulting equation, S = T, for the unknown. Example 4.7.1: Solving an Exponential Equation with a Common Base. Solve 2x − 1 = 22x − 4.

  16. The 11 Natural Log Rules You Need to Know

    Natural logs may seem difficult, but once you understand a few key natural log rules, you'll be able to easily solve even very complicated-looking problems.

  17. Lesson Explainer: Natural Logarithmic Equations

    In this explainer, we will learn how to use natural logarithms to solve exponential and logarithmic equations. Recall that a logarithm is simply the power a number must be raised to in order to give a certain value. For example, the logarithmic equation l o g 𝑏 = 𝑐 is another way of writing the exponential equation 𝑎 = 𝑏 .

  18. How to Solve Logarithmic Equations in Algebra

    Thus, we have to solve the following log equations: log x = 2 . log x = 3 . The obvious solutions are. x_1 = 10^2 = 100 ,, \quad \quad x_2 = 10^3 = 1000 . Example 3. log_x 3 log_{ 3x } 3 = log_{ 9x } 3 . Although this log equation seems quite different from those we have considered so far, we will now show how it can be solved using the same ...

  19. How to evaluate logs using the general log rule

    But log problems can get a little more complicated than this, and that's what we want to talk about here. Hi! I'm krista. I create online courses to help you rock your math class. ... This method for solving logs will always work. If we can get the bases equal to one another, then we can also set the exponents equal to each other. ...

  20. How solve complicated logarithm?

    How solve complicated logarithm? | Socratic How solve complicated logarithm? Can't someone please explain to me how to do question 6e and 10b? Thank! Precalculus 1 Answer Manikandan S. Oct 3, 2017 Use definitions Explanation: 6 e ) log3(x2 −3x −1) = 0 Take 3f both sides. You get 3log3(x2−3x−1) = 30 x2 −3x − 1 = 1 Now solve the quadratic equation

  21. 2.2 Complex Logarithms

    Principal Log. Remarking that the complex exponential function, , is defined in terms of the radial part, , and an angular part , it is natural that the complex logarithm be defined in terms of the real natural logarithm of a modulus, and an argument function. For , we define the principal branch of the complex logarithm by where is the ...

  22. Solve your critical errors

    Open the Log details window. Looking at our example, the APM agent has pulled logs related to api-gateway, giving us event data about our error: Our logs-in-context capability formats your logs information, though you have the option to investigate the unformatted log as well.

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    Why it matters: The previously unreported meeting, recounted to Axios by three people familiar with the events, is emblematic of the Biden administration's struggle with the border crisis during the past three years — infighting, blame-shifting and indecision. Biden's fury subsided, and aides scrambled for the information he wanted.

  25. Intro to Logarithms (article)

    The answer would be 4 . This is expressed by the logarithmic equation log 2 ( 16) = 4 , read as "log base two of sixteen is four". 2 4 = 16 log 2 ( 16) = 4. Both equations describe the same relationship between the numbers 2 , 4 , and 16 , where 2 is the base and 4 is the exponent. The difference is that while the exponential form isolates the ...

  26. Inside the race to grasp the fate of the Colorado River

    Over the years, Reclamation has looked to climate science to solve the real world problem of a shrinking Colorado River, according to river experts and people involved in the effort.

  27. Solving Logarithmic Equations

    46K 3.3M views 6 years ago New Algebra Playlist This algebra video tutorial explains how to solve logarithmic equations with logs on both sides. It explains how to convert from logarithmic form...