Solving Logarithmic Equations

Show your work for each problem.

1.
Which equation is the exponential form of $\log_4 x = 3$ ?
1. a) $x = 4^3$
2. b) $x = 3^4$
3. c) $x = \frac{4}{3}$
4. d) $x = 3 \cdot 4$
Answer: $x = 4^3$
$\log_b A = c \iff A = b^c$ . Here $b = 4$ and $c = 3$ , so $x = 4^3 = 64$ .
2.
Solve $\log_5 x = 3$ .
Answer: 125
$x = 5^3 = 125$ . Check: $\log_5 125 = 3$ because $5^3 = 125$ .
3.
Solve $\log_4(x - 3) = 2$ .
Answer: 19
1. Rewrite in exponential form: $x - 3 = 4^2$ . What is $4^2$ ? 16
2. Solve $x - 3 = 16$ . What is $x$ ? 19
3. Domain check: what is the argument $x - 3$ for your $x$ ? (It must be positive.) 16
4.
Which value can NOT be a solution of an equation that contains $\log_3(x - 5)$ ?
1. a) $x = 32$
2. b) $x = 14$
3. c) $x = 4$
4. d) $x = 6$
Answer: $x = 4$
The argument must be positive: $x - 5 > 0$ , so $x > 5$ . For $x = 4$ the argument is $4 - 5 = -1$ , and $\log_3(-1)$ is not defined.
5.
Solve $\log_2(3x - 1) = 3$ .
Answer: 3
$3x - 1 = 2^3 = 8$ , so $3x = 9$ and $x = 3$ . Check: the argument $3 \cdot 3 - 1 = 8$ is positive and $\log_2 8 = 3$ .
6.
Solve $\ln x = 2$ . Round to two decimal places.
Answer: 7.39
1. Rewrite in exponential form: $x = e^{\square}$ . What is the exponent? 2
2. Calculate $e^2$ and round to two decimal places. 7.39
7.
Solve $\log_3(5x - 4) = \log_3(2x + 11)$ .
Answer: 5
1. Set the arguments equal: $5x - 4 = 2x + 11$ . Collect terms: $3x = \square$ 15
2. What is $x$ ? 5
3. Check: what is the argument $5x - 4$ for this $x$ ? 21
8.
Solve $2 \log_5 x = \log_5 36$ .
Answer: 6
Power rule: $\log_5(x^2) = \log_5 36$ , so $x^2 = 36$ and $x = 6$ or $x = -6$ . In the original equation $\log_5(-6)$ is not defined, so $x = -6$ is extraneous. The solution is $x = 6$ .
9.
Which expression is the exact solution of $7^x = 50$ ?
1. a) $\log 50 - \log 7$
2. b) $\frac{\log 50}{\log 7}$
3. c) $\frac{50}{7}$
4. d) $\log \frac{50}{7}$
Answer: $\frac{\log 50}{\log 7}$
$\log(7^x) = \log 50$ gives $x \log 7 = \log 50$ , so $x = \frac{\log 50}{\log 7} \approx 2.01$ . Note that $\log \frac{50}{7}$ and $\log 50 - \log 7$ are the same number (about $0.85$ ) and are wrong.
10.
Solve $5^x = 40$ by taking the common logarithm (base 10) of both sides. Round to two decimal places.
Answer: 2.29
1. After the power rule you have $x \log 5 = \log 40$ . What is $\log 40$ , rounded to three decimal places? 1.602
2. What is $\log 5$ , rounded to three decimal places? 0.699
3. Divide: $x = \frac{\log 40}{\log 5}$ . Round $x$ to two decimal places. 2.29
11.
Solve $2^x = 20$ . Round to two decimal places.
Answer: 4.32
$x \log 2 = \log 20$ , so $x = \frac{\log 20}{\log 2} \approx 4.32$ . This fits the estimate $2^4 = 16 < 20 < 32 = 2^5$ .
12.
Solve $\log_2(x + 6) - \log_2(x - 1) = 3$ .
Answer: 2
1. Condense: $\log_2 \frac{x + 6}{x - 1} = 3$ . Rewrite in exponential form: $\frac{x + 6}{x - 1} = \square$ 8
2. Multiply by $(x - 1)$ : $x + 6 = 8x - 8$ . Collect terms: $7x = \square$ 14
3. What is $x$ ? (Check that $x - 1$ is positive.) 2
13.
Solve $\log_6 x + \log_6(x + 1) = 1$ .
Answer: 2
1. Condense: $\log_6\big(x(x + 1)\big) = 1$ . Exponential form: $x(x + 1) = \square$ 6
2. $x^2 + x - 6 = (x + 3)(x - 2) = 0$ . What is the positive candidate? 2
3. What is the other candidate? -3
4. Which candidate is the solution? (All arguments must be positive.) 2
14.
Solve $\log_2 x + \log_2(x - 6) = 4$ .
1. a)only $x = 8$
2. b) $x = 8$ and $x = -2$
3. c)no solution
4. d)only $x = -2$
Answer: only $x = 8$
$\log_2\big(x(x - 6)\big) = 4$ , so $x^2 - 6x = 16$ , $x^2 - 6x - 16 = 0$ , $(x - 8)(x + 2) = 0$ . The candidate $x = -2$ makes $\log_2(-2)$ undefined, so only $x = 8$ remains: $\log_2 8 + \log_2 2 = 3 + 1 = 4$ ✓
15.
Solve $\log_3 x + \log_3(x - 8) = 2$ .
Answer: 9
1. Condense and rewrite: $x(x - 8) = 3^2 = \square$ 9
2. $x^2 - 8x - 9 = (x - 9)(x + 1) = 0$ . What is the positive candidate? 9
3. What is the other candidate? -1
4. Which candidate is the solution? (Check the arguments $x$ and $x - 8$ .) 9
16.
Solve $\log(x + 9) - \log x = 1$ , where $\log$ is the common logarithm (base 10).
Answer: 1
$\log \frac{x + 9}{x} = 1$ , so $\frac{x + 9}{x} = 10$ , $x + 9 = 10x$ , $9x = 9$ , $x = 1$ . Check: $\log 10 - \log 1 = 1 - 0 = 1$ ✓
17.
Solve $\log(x + 3) + \log x = 1$ (common logarithm, base 10).
1. a)only $x = 2$
2. b) $x = 2$ and $x = -5$
3. c)no solution
4. d)only $x = -5$
Answer: only $x = 2$
$\log\big(x(x + 3)\big) = 1$ , so $x^2 + 3x = 10$ , $(x + 5)(x - 2) = 0$ . For $x = -5$ both $\log(-2)$ and $\log(-5)$ are undefined, so it is extraneous. The solution is $x = 2$ : $\log 5 + \log 2 = \log 10 = 1$ ✓
18.
Solve $4 \cdot 3^{x - 1} = 100$ . Round to two decimal places.
Answer: 3.93
1. Divide both sides by 4. $3^{x - 1} = \square$ 25
2. Take logarithms: $x - 1 = \frac{\log 25}{\log 3}$ . Round $x - 1$ to two decimal places. 2.93
3. What is $x$ , rounded to two decimal places? 3.93