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8.11: Solving Logarithmic Equations

Difficulty Level: At Grade Created by: CK-12
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"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*}." What number are you thinking of?

Guidance

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*}, to cancel out the log.

Example A

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

Solution: 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*}

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*}

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*}

Example B

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

Solution: 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*}

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*} 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*}

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*}

Example C

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

Solution: 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*}

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

\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*}

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*}

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

Intro Problem Revisit We can rewrite \begin{align*}\log 10x^2 - \log x = 3\end{align*} as \begin{align*}\log {\frac{10x^2}{x}} = 3\end{align*} 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*}

Therefore, the number you are thinking of is 100.

Guided Practice

Solve the following logarithmic equations.

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

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

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

Answers

1. 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*}

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

\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*}

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

3. 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*}

Practice

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

  1. \begin{align*}\log_2 x =15\end{align*}
  2. \begin{align*}\log x = 125\end{align*}
  3. \begin{align*}\log_9 (x-5) =16\end{align*}
  4. \begin{align*}\log_7(2x+3)=3\end{align*}
  5. \begin{align*}8 \ln(3-x)=5\end{align*}
  6. \begin{align*}4 \log_3 3x-\log_3 x=5\end{align*}
  7. \begin{align*}\log(x+5) + \log x = \log 14\end{align*}
  8. \begin{align*}2 \ln x - \ln x =0\end{align*}
  9. \begin{align*}3 \log_3(x-5) = 3\end{align*}
  10. \begin{align*}\frac{2}{3} \log_3 x=2\end{align*}
  11. \begin{align*}5 \log \frac{x}{2} -3 \log \frac{1}{x} = \log 8\end{align*}
  12. \begin{align*}2 \ln x^{e+2} - \ln x=10\end{align*}
  13. \begin{align*}2 \log_6 x+1 = \log_6(5x+4)\end{align*}
  14. \begin{align*}2 \log_{\frac{1}{2}}x+2=\log_{\frac{1}{2}}(x+10)\end{align*}
  15. \begin{align*}3 \log_{\frac{2}{3}} x-\log_{\frac{2}{3}} 27 = \log_{\frac{2}{3}}8\end{align*}

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Mar 12, 2013
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