<meta http-equiv="refresh" content="1; url=/nojavascript/"> Solving Logarithmic Equations ( Read ) | Analysis | CK-12 Foundation
Dismiss
Skip Navigation

Solving Logarithmic Equations

%
Best Score
Practice Solving Logarithmic Equations
Practice
Best Score
%
Practice Now
Solving Logarithmic Equations
 0  0  0

"I'm thinking of another number," you tell your best friend. "The number I'm thinking of satisfies the equation \log 10x^2 - \log x = 3 ." 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, b^{\log_b x}=x , to cancel out the log.

Example A

Solve \log_2(x+5)=9 .

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

\log_2(x+5) &= 9 \\2^9 &= x+5 \\512 &= x+5 \\507 &= x

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

2^{\log_2(x+5)} &= 2^9 \\x+5 &= 512 \\x &= 507

Make sure to check your answers for logarithmic equations. There can be times when you get an extraneous solution. \log_2(507+5)=9 \rightarrow \log_2 512=9

Example B

Solve 3 \ln(-x)-5=10 .

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

3 \ln(-x)-5 &= 10 \\3 \ln(-x) &= 15 \\\ln(-x)&= 5

Recall that the inverse of the natural log is the natural number. Therefore, everything needs to be put into the exponent of e in order to get rid of the log.

e^{\ln(-x)} &= e^5 \\-x &= e^5 \\x &= -e^5 \approx -148.41

Checking the answer, we have 3 \ln(-(-e^5))-5=10 \rightarrow 3\ln e^5 -5 =10 \rightarrow 3 \cdot 5-5=10

Example C

Solve \log 5x + \log(x-1)=2

Solution: Condense the left-hand side using the Product Property.

\log 5x + \log (x-1)=2 \\\log [5x(x-1)]=2 \\\log (5x^2-5x)=2

Now, put everything in the exponent of 10 and solve for x .

10^{\log(5x^2-5x)} &= 10^2 \\5x^2 - 5x &= 100 \\x^2-x-20 &= 0 \\(x-5)(x+4) &= 0 \\x &=5, -4

Now, check both answers.

\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

-4 is an extraneous solution. In the step \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.

Intro Problem Revisit We can rewrite \log 10x^2 - \log x = 3 as \log {\frac{10x^2}{x}} = 3 and solve for x .

\log {\frac{10x^2}{x}} = 3\\\log 10x = 3\\10^{\log10x} = 10^3\\10x = 1000\\x = 100

Therefore, the number you are thinking of is 100.

Guided Practice

Solve the following logarithmic equations.

1. 9 + 2 \log_3 x=23

2. \ln (x-1)-\ln(x+1)=8

3. \frac{1}{2}\log_5(2x+5)=5

Answers

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

9 + 2 \log_3 x &= 23 \\2 \log_3 x &= 14 \\\log_3 x &= 7 \\x &= 3^7=2187

2. Condense the left-hand side using the Quotient Rule and put everything in the exponent of e .

\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

Checking our answer, we get \ln (-2.85-1) - \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.

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

\frac{1}{2} \log_5(2x+5)&= 2 \\\log_5(2x+5)&=4 \\2x+5 &= 625 \\2x &=620  \\x &= 310

Explore More

Use properties of logarithms and a calculator to solve the following equations for x . Round answers to three decimal places and check for extraneous solutions.

  1. \log_2 x =15
  2. \log_{12} x = 2.5
  3. \log_9 (x-5) =2
  4. \log_7(2x+3)=3
  5. 8 \ln(3-x)=5
  6. 4 \log_3 3x-\log_3 x=5
  7. \log(x+5) + \log x = \log 14
  8. 2 \ln x - \ln x =0
  9. 3 \log_3(x-5) = 3
  10. \frac{2}{3} \log_3 x=2
  11. 5 \log \frac{x}{2} -3 \log \frac{1}{x} = \log 8
  12. 2 \ln x^{e+2} - \ln x=10
  13. 2 \log_6 x+1 = \log_6(5x+4)
  14. 2 \log_{\frac{1}{2}}x+2=\log_{\frac{1}{2}}(x+10)
  15. 3 \log_{\frac{2}{3}} x-\log_{\frac{2}{3}} 27 = \log_{\frac{2}{3}}8

Image Attributions

Reviews

Email Verified
Well done! You've successfully verified the email address .
OK
Please wait...
Please wait...
ShareThis Copy and Paste

Original text