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# Solving Logarithmic Equations

## Use technology to solve equations with logs

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Practice Solving Logarithmic Equations
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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 $\log 10x^2 - \log x = 3$ ." 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 $x$ .

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

$\log 5(5) + \log(5-1) &= 2 \qquad \qquad \log5(-4) + \log((-4)-1)= 2 \\\log 25 + \log 4 &= 2 \qquad \qquad \quad \ \log(-20) + \log(-5) = 2 \\\log 100 &= 2$

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

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.

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

### Practice

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. $3 \log_{\frac{2}{3}} x-\log_{\frac{2}{3}} 27 = \log_{\frac{2}{3}}8$