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Solving Logarithmic Equations

Use technology to solve equations with logs

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Solving Logarithmic Equations

"I'm thinking of another number," you tell your best friend. "The number I'm thinking of satisfies the equation \begin{align*}\log 10x^2 - \log x = 3\end{align*}log10x2logx=3." What number are you thinking of?

Solving Logarithm Equations

A logarithmic equation has the variable within the log. To solve a logarithmic equation, you will need to use the inverse property, \begin{align*}b^{\log_b x}=x\end{align*}blogbx=x, to cancel out the log.

Solve the following problems

Solve \begin{align*}\log_2(x+5)=9\end{align*}log2(x+5)=9.

There are two different ways to solve this equation. The first is to use the definition of a logarithm.

\begin{align*}\log_2(x+5) &= 9 \\ 2^9 &= x+5 \\ 512 &= x+5 \\ 507 &= x\end{align*}log2(x+5)29512507=9=x+5=x+5=x

The second way to solve this equation is to put everything into the exponent of a 2, and then use the inverse property.

\begin{align*}2^{\log_2(x+5)} &= 2^9 \\ x+5 &= 512 \\ x &= 507\end{align*}2log2(x+5)x+5x=29=512=507

Make sure to check your answers for logarithmic equations. There can be times when you get an extraneous solution. \begin{align*}\log_2(507+5)=9 \rightarrow \log_2 512=9 \end{align*}log2(507+5)=9log2512=9

Solve \begin{align*}3 \ln(-x)-5=10\end{align*}3ln(x)5=10.

First, add 5 to both sides and then divide by 3 to isolate the natural log.

\begin{align*}3 \ln(-x)-5 &= 10 \\ 3 \ln(-x) &= 15 \\ \ln(-x)&= 5\end{align*}3ln(x)53ln(x)ln(x)=10=15=5

Recall that the inverse of the natural log is the natural number. Therefore, everything needs to be put into the exponent of \begin{align*}e\end{align*}e in order to get rid of the log.

\begin{align*}e^{\ln(-x)} &= e^5 \\ -x &= e^5 \\ x &= -e^5 \approx -148.41\end{align*}eln(x)xx=e5=e5=e5148.41

Checking the answer, we have \begin{align*}3 \ln(-(-e^5))-5=10 \rightarrow 3\ln e^5 -5 =10 \rightarrow 3 \cdot 5-5=10\end{align*}3ln((e5))5=103lne55=10355=10

Solve \begin{align*}\log 5x + \log(x-1)=2\end{align*}log5x+log(x1)=2

Condense the left-hand side using the Product Property.

\begin{align*}\log 5x + \log (x-1)=2 \\ \log [5x(x-1)]=2 \\ \log (5x^2-5x)=2\end{align*}log5x+log(x1)=2log[5x(x1)]=2log(5x25x)=2

Now, put everything in the exponent of 10 and solve for \begin{align*}x\end{align*}x.

\begin{align*}10^{\log(5x^2-5x)} &= 10^2 \\ 5x^2 - 5x &= 100 \\ x^2-x-20 &= 0 \\ (x-5)(x+4) &= 0 \\ x &=5, -4\end{align*}10log(5x25x)5x25xx2x20(x5)(x+4)x=102=100=0=0=5,4

Now, check both answers.

\begin{align*}\log 5(5) + \log(5-1) &= 2 \qquad \qquad \log5(-4) + \log((-4)-1)= 2 \\ \log 25 + \log 4 &= 2 \qquad \qquad \quad \ \log(-20) + \log(-5) = 2 \\ \log 100 &= 2\end{align*}log5(5)+log(51)log25+log4log100=2log5(4)+log((4)1)=2=2 log(20)+log(5)=2=2

-4 is an extraneous solution. In the step \begin{align*}\log(-20) + \log(-5)=2\end{align*}log(20)+log(5)=2, we cannot take the log of a negative number, therefore -4 is not a solution. 5 is the only solution.

Examples

Example 1

Earlier, you were asked what number are you thinking of. 

We can rewrite \begin{align*}\log 10x^2 - \log x = 3\end{align*}log10x2logx=3 as \begin{align*}\log {\frac{10x^2}{x}} = 3\end{align*}log10x2x=3 and solve for x.

\begin{align*}\log {\frac{10x^2}{x}} = 3\\ \log 10x = 3\\ 10^{\log10x} = 10^3\\ 10x = 1000\\ x = 100\end{align*}log10x2x=3log10x=310log10x=10310x=1000x=100

Therefore, the number you are thinking of is 100.

Solve the following logarithmic equations.

Example 2

\begin{align*}9 + 2 \log_3 x=23\end{align*}9+2log3x=23

Isolate the log and put everything in the exponent of 3.

\begin{align*}9 + 2 \log_3 x &= 23 \\ 2 \log_3 x &= 14 \\ \log_3 x &= 7 \\ x &= 3^7=2187\end{align*}9+2log3x2log3xlog3xx=23=14=7=37=2187

Example 3

\begin{align*}\ln (x-1)-\ln(x+1)=8\end{align*}ln(x1)ln(x+1)=8

Condense the left-hand side using the Quotient Rule and put everything in the exponent of \begin{align*}e\end{align*}e.

\begin{align*}\ln(x-1) - \ln(x+1) &=8 \\ \ln \left(\frac{x-1}{x+1}\right) &= 8 \\ \frac{x-1}{x+1} &= \ln 8 \\ x-1 &=(x+1) \ln 8 \\ x-1 &= x \ln 8 + \ln 8 \\ x-x \ln 8 &= 1 + \ln 8 \\ x(1- \ln 8) &= 1 + \ln 8 \\ x &= \frac{1+ \ln 8}{1- \ln 8} \approx -2.85\end{align*}ln(x1)ln(x+1)ln(x1x+1)x1x+1x1x1xxln8x(1ln8)x=8=8=ln8=(x+1)ln8=xln8+ln8=1+ln8=1+ln8=1+ln81ln82.85

Checking our answer, we get \begin{align*}\ln (-2.85-1) - \ln (2.85+1)=8\end{align*}ln(2.851)ln(2.85+1)=8, which does not work because the first natural log is of a negative number. Therefore, there is no solution for this equation.

Example 4

\begin{align*}\frac{1}{2}\log_5(2x+5)=5\end{align*}12log5(2x+5)=5

 Multiply both sides by 2 and put everything in the exponent of a 5.

\begin{align*}\frac{1}{2} \log_5(2x+5)&= 2 \\ \log_5(2x+5)&=4 \\ 2x+5 &= 625 \\ 2x &=620 \\ x &= 310\end{align*}12log5(2x+5)log5(2x+5)2x+52xx=2=4=625=620=310

Review

Use properties of logarithms and a calculator to solve the following equations for \begin{align*}x\end{align*}x. Round answers to three decimal places and check for extraneous solutions.

  1. \begin{align*}\log_2 x =15\end{align*}log2x=15
  2. \begin{align*}\log_{12} x = 2.5\end{align*}log12x=2.5
  3. \begin{align*}\log_9 (x-5) =2\end{align*}log9(x5)=2
  4. \begin{align*}\log_7(2x+3)=3\end{align*}log7(2x+3)=3
  5. \begin{align*}8 \ln(3-x)=5\end{align*}8ln(3x)=5
  6. \begin{align*}4 \log_3 3x-\log_3 x=5\end{align*}4log33xlog3x=5
  7. \begin{align*}\log(x+5) + \log x = \log 14\end{align*}log(x+5)+logx=log14
  8. \begin{align*}2 \ln x - \ln x =0\end{align*}2lnxlnx=0
  9. \begin{align*}3 \log_3(x-5) = 3\end{align*}3log3(x5)=3
  10. \begin{align*}\frac{2}{3} \log_3 x=2\end{align*}23log3x=2
  11. \begin{align*}5 \log \frac{x}{2} -3 \log \frac{1}{x} = \log 8\end{align*}5logx23log1x=log8
  12. \begin{align*}2 \ln x^{e+2} - \ln x=10\end{align*}2lnxe+2lnx=10
  13. \begin{align*}2 \log_6 x+1 = \log_6(5x+4)\end{align*}2log6x+1=log6(5x+4)
  14. \begin{align*}2 \log_{\frac{1}{2}}x+2=\log_{\frac{1}{2}}(x+10)\end{align*}2log12x+2=log12(x+10)
  15. \begin{align*}3 \log_{\frac{2}{3}} x-\log_{\frac{2}{3}} 27 = \log_{\frac{2}{3}}8\end{align*}3log23xlog2327=log238

Answers for Review Problems

To see the Review answers, open this PDF file and look for section 8.11. 

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