Practice 1 Solutions

The approximate value of the log is between which two integers?

$\log 154$

$10^{x} = 154 \begin{matrix} 10^{2} = 100 & 10^{3} = 1000 \end{matrix}$

Between 2 and 3

Note

Q: To which integer is log 154 closer? Explain.

A: Log 154 is closer to 2 since 154 is closer to 100 than 1,000.

$\log 4$

$10^{x} = 4 \begin{matrix} 10^{0} = 1 & 10^{1} = 10 \end{matrix}$

Between 0 and 1

$\log 6438$

$10^{x} = 6438 \begin{matrix} 10^{3} = 1000 & 10^{4} = 10000 \end{matrix}$

Between 3 and 4

$\log \frac{1}{25}$

$10^{x} = \frac{1}{25} \begin{matrix} 10^{- 1} = \frac{1}{10} & 10^{- 2} = \frac{1}{100} \end{matrix}$

Between –2 and –1

For problems 5–8:

1. Write the log in terms of X and Y when $\log 2 = X$ and $\log 3 = Y$ .
2. Calculate the approximate value of the log when $\log 2 \approx 0.301$ and $\log 3 \approx 0.4771$ .

$\log 108$

$\begin{array}{rcl} \begin{matrix} \log 108 = \log (2^{2} \cdot 3^{3}) \\ = \log 2^{2} + \log 3^{3} \end{matrix} \end{array}$

$\begin{matrix} = 2 \log 2 + 3 \log 3 \\ = 2 (X) + 3 (Y) \\ = 2 X + 3 Y \end{matrix}$
$\begin{matrix} \approx 2 (0.301) + 3 (0.4771) \\ \approx 2.0333 \end{matrix}$

$2 X + 3 Y$
2.0333

$\log 96$

$\begin{array}{rcl} \begin{matrix} \log 96 = \log (2^{5} \cdot 3) \\ = \log 2^{5} + \log 3 \end{matrix} \end{array}$

$\begin{matrix} = 5 \log 2 + \log 3 \\ = 5 (X) + (Y) \\ = 5 X + Y \end{matrix}$
$\begin{matrix} \approx 5 (0.301) + (0.4771) \\ \approx 1.9821 \end{matrix}$

$5 X + Y$
1.9821

$\log 162$

$\begin{array}{rcl} \log 162 & = & \log (2 \cdot 3^{4}) \\ = & \log 2 + \log 3^{4} \end{array}$

$\begin{matrix} = \log 2 + 4 \log 3 \\ = (X) + 4 (Y) \\ = X + 4 Y \end{matrix}$
$\begin{matrix} \approx (0.301) + 4 (0.4771) \\ \approx 2.2094 \end{matrix}$

$X + 4 Y$
2.2094

$\log 144$

$\begin{array}{rcl} \log 144 & = & \log (2^{4} \cdot 3^{2}) \\ = & \log 2^{4} + \log 3^{2} \end{array}$

$\begin{matrix} = 4 \log 2 + 2 \log 3 \\ = 4 (X) + 2 (Y) \\ = 4 X + 2 Y \end{matrix}$
$\begin{matrix} \approx 4 (0.301) + 2 (0.4771) \\ \approx 2.1582 \end{matrix}$

$4 X + 2 Y$
2.1582

Solve. Write the answer as a logarithm and as a number to four decimal places.

$6^{x} = 30$

$\begin{array}{rcl} \log 6^{x} & = & \log 30 \\ x \log 6 & = & \log 30 \end{array}$

$x = \frac{\log 30}{\log 6} \approx 1.8982$

$81^{x - 3} = 15$

$\begin{array}{rcl} \log 81^{x - 3} & = & \log 15 \\ (x - 3) \log 81 & = & \log 15 \\ x - 3 & = & \frac{\log 15}{\log 81} \end{array}$

$x = \frac{\log 15}{\log 81} + 3 \approx 3.6162$

Note

When the power contains an expression, use parentheses to ensure you correctly complete the steps of solving for the variable.

$25^{3 x + 1} = 520$

$\begin{array}{rcl} \log 25^{3 x + 1} & = & \log 520 \\ (3 x + 1) \log 25 & = & \log 520 \\ 3 x + 1 & = & \frac{\log 520}{\log 25} \\ 3 x & = & \frac{\log 520}{\log 25} - 1 \end{array}$

$x = \frac{\log 520}{3 \log 25} - \frac{1}{3} \approx 0.3143$

Note

Remember to divide every term by 3 when solving for x.

$7^{2 x} = 490$

$\begin{array}{rcl} \log 7^{2 x} & = & \log 490 \\ 2 x \log 7 & = & \log 490 \\ 2 x & = & \frac{\log 490}{\log 7} \end{array}$

$x = \frac{\log 490}{2 \log 7} \approx 1.5916$

$2^{x} = 56$

$\begin{array}{rcl} \log 2^{x} & = & \log 56 \\ x \log 2 & = & \log 56 \end{array}$

$x = \frac{\log 56}{\log 2} \approx 5.8074$

$32^{x + 1} = 12$

$\begin{array}{rcl} \log 32^{x + 1} & = & \log 12 \\ (x + 1) \log 32 & = & \log 12 \\ x + 1 & = & \frac{\log 12}{\log 32} \end{array}$

$x = \frac{\log 12}{\log 32} - 1 \approx - 0.2830$

Write with common logs using the Change of Base Rule.

Note

Problems 15–17

If you need more practice, you can also evaluate these expressions using technology. Because the directions do not tell you to round, it is not necessary to use 4 decimal places.

$\log_{5} 9^{4}$

$\frac{\log {(9)}^{4}}{\log 5}$

$\frac{4 \log 9}{\log 5}$

$\log_{3} 11 x$

$\frac{\log 11 x}{\log 3}$

$\frac{\log 11 + \log x}{\log 3}$

$\log_{2} 6 + \log_{3} 4$

$\frac{\log 6}{\log 2} + \frac{\log 4}{\log 3}$

Solve. Write as a common log.

$7^{\frac{x + 2}{3}} = 84$

$\begin{array}{rcl} \log 7^{\frac{x + 2}{3}} & = & \log 84 \\ (\frac{x + 2}{3}) \log 7 & = & \log 84 \\ (3) \frac{x + 2}{3} & = & \frac{\log 84}{\log 7} (3) \\ x + 2 & = & \frac{3 \log 84}{\log 7} \end{array}$

$x = \frac{3 \log 84}{\log 7} - 2$

$2^{4 - x} = 13$

$\begin{array}{rcl} \log 2^{4 - x} & = & \log 13 \\ (4 - x) \log 2 & = & \log 13 \\ 4 - x & = & \frac{\log 13}{\log 2} \\ (- 1) (- x) & = & (\frac{\log 13}{\log 2} - 4) (- 1) \end{array}$

$x = - \frac{\log 13}{\log 2} + 4$

$15^{\frac{x}{7}} = 25$

$\begin{array}{rcl} \log 15^{\frac{x}{7}} & = & \log 25 \\ \frac{x}{7} \log 15 & = & \log 25 \\ (7) \frac{x}{7} & = & \frac{\log 25}{\log 15} (7) \end{array}$

$x = \frac{7 \log 25}{\log 15}$