Practice 2 Solutions

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

$\log 890$

$10^{x} = 890 10^{2} = 100, 10^{3} = 1, 000$

Between 2 and 3

$\log \frac{7}{500}$

$10^{x} = \frac{7}{500} \frac{7}{500} = \frac{14}{1000} 10^{- 2} = \frac{1}{100}, 10^{- 3} = \frac{1}{1000}$

Between –2 and –3

$\log 0.2$

$10^{x} = 0.2 10^{x} = \frac{2}{10} 10^{- 1} = \frac{1}{10}, 10^{- 2} = \frac{1}{100}$

Between –1 and –2

$\log 83$

$10^{x} = 83 10^{1} = 10, 10^{2} = 100$

Between 1 and 2

For problems 5–8:

1. Write the log in terms of X and Y when $\log 2 = X$ and $\log 5 = Y$ .
2. Calculate the approximate value of the log when $\log 2 \approx 0.301$ and $\log 5 \approx 0.6990$ .

$\log 20$

$\begin{array}{rcl} \log 20 & = & \log (2^{2} \cdot 5) \\ = & \log 2^{2} + \log 5 \end{array}$

$\begin{matrix} = 2 \log 2 + \log 5 \\ = 2 (X) + (Y) \\ = 2 X + Y \end{matrix}$
$\begin{matrix} \approx 2 (0.301) + (0.6990) \\ \approx 1.301 \end{matrix}$

$2 X + Y$
1.301

$\log 200$

$\begin{array}{rcl} \log 200 & = & \log (2^{3} \cdot 5^{2}) \\ = & \log 2^{3} + \log 5^{2} \end{array}$

$\begin{matrix} = 3 \log 2 + 2 \log 5 \\ = 3 (X) + 2 (Y) \\ = 3 X + 2 Y \end{matrix}$
$\begin{matrix} \approx 3 (0.301) + 2 (0.6990) \\ \approx 2.301 \end{matrix}$

$3 X + 2 Y$
2.301

Note

Alternate solving method

$\log 100 = 2$ $\begin{array}{rcl} \log 200 & = & \log (100 \cdot 2) \\ = & \log 100 + \log 2 \\ = & 2 + 0.301 \\ = & 2.301 \end{array}$ $\log 100 = 2$

$\log 250$

$\begin{array}{rcl} \log 250 & = & \log (2 \cdot 5^{3}) \\ = & \log 2 + \log 5^{3} \end{array}$

$\begin{matrix} = \log 2 + 3 \log 5 \\ = (X) + 3 (Y) \\ = X + 3 Y \end{matrix}$
$\begin{matrix} \approx (0.301) + 3 (0.6990) \\ \approx 2.398 \end{matrix}$

$X + 3 Y$
2.398

$\log \frac{8}{25}$

$\begin{array}{rcl} \log \frac{8}{25} & = & \log (\frac{2^{3}}{5^{2}}) \\ = & \log 2^{3} - \log 5^{2} \end{array}$

$\begin{matrix} = 3 \log 2 - 2 \log 5 \\ = 3 (X) - 2 (Y) \\ = 3 X - 2 Y \end{matrix}$
$\begin{matrix} \approx 3 (0.301) - 2 (0.6990) \\ \approx - 0.495 \end{matrix}$

$3 X - 2 Y$
–0.495

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

$17^{x + 1} = 34$

$\begin{array}{rcl} \log 17^{x + 1} & = & \log 34 \\ (x + 1) \log 17 & = & \log 34 \\ x + 1 & = & \frac{\log 34}{\log 17} \end{array}$

$x = \frac{\log 34}{\log 17} - 1 \approx 0.2447$

$2^{x} = 99$

$\begin{array}{rcl} \log 2^{x} & = & \log 99 \\ x \log 2 & = & \log 99 \end{array}$

$x = \frac{\log 99}{\log 2} \approx 6.6293$

$6^{\frac{x}{4} + 2} = 51$

$\begin{array}{rcl} \log 6^{\frac{x}{4} + 2} & = & \log 51 \\ (\frac{x}{4} + 2) \log 6 & = & \log 51 \\ \frac{x}{4} + 2 & = & \frac{\log 51}{\log 6} \\ (4) \frac{x}{4} & = & (\frac{\log 51}{\log 6} - 2) (4) \end{array}$

$x = \frac{4 \log 51}{\log 6} - 8 \approx 0.7776$

$54^{2 x} = 100$

$\begin{array}{rcl} \log 54^{2 x} & = & \log 100 \\ 2 x \log 54 & = & 2 \\ 2 x & = & \frac{2}{\log 54} \\ x & = & \frac{2}{2 \log 54} \end{array}$

$x = \frac{1}{\log 54} \approx 0.5772$

Note

Recall log 100 = 2.

Q: What does $\log 100$ equal? Explain.

A: log 100 = 2 because $10^{2} = 100$

$3^{1 - x} = 10$

$\begin{array}{rcl} \log 3^{1 - x} & = & \log 10 \\ (1 - x) \log 3 & = & 1 \\ 1 - x & = & \frac{1}{\log 3} \\ - x & = & \frac{1}{\log 3} - 1 \end{array}$

$x = - \frac{1}{\log 3} + 1 \approx - 1.0959$

Note

Recall log 10 = 1

$7^{5 x + 4} = 56$

$\begin{array}{rcl} \log 7^{5 x + 4} & = & \log 56 \\ (5 x + 4) \log 7 & = & \log 56 \\ 5 x + 4 & = & \frac{\log 56}{\log 7} \\ (\frac{1}{5}) 5 x & = & (\frac{\log 56}{\log 7} - 4) (\frac{1}{5}) \end{array}$

$x = \frac{\log 56}{5 \log 7} - \frac{4}{5} \approx - 0.3863$

Write with common logs using the Change of Base Rule.

$\log_{m} 8 x$

$\begin{array}{rcl} \log_{m} 8 x & = & \log_{m} (8 \cdot x) \\ = & \log_{m} 8 + \log_{m} x \end{array}$

$\begin{array}{rcl} \frac{\log 8}{\log m} + \frac{\log x}{\log m} \end{array}$

$\log_{25} 50$

$\begin{array}{rcl} \log_{25} 50 & = & \log_{25} (25 \cdot 2) \\ = & \log_{25} 25 + \log_{25} 2 \end{array}$

$\begin{array}{rcl} 1 + \frac{\log 2}{\log 25} \end{array}$

Note

Another way to write the solution is $\frac{\log 50}{\log 25}$ .

Recall $\log_{x} x = 1$

$\log_{b} r$

$\log_{b} r$

$\frac{\log r}{\log b}$

Solve. Write as a common log.

$13^{\frac{x}{2}} = 7$

$\begin{array}{rcl} \log 13^{\frac{x}{2}} & = & \log 7 \\ \frac{x}{2} \log 13 & = & \log 7 \\ (2) \frac{x}{2} & = & \frac{\log 7}{\log 13} (2) \end{array}$

$x = \frac{2 \log 7}{\log 13}$ $x = \frac{2 \log 7}{\log 13}$

$8^{\frac{2 x}{3}} = 100$

$\begin{array}{rcl} \log 8^{\frac{2 x}{3}} & = & \log 100 \\ \frac{2 x}{3} \log 8 & = & 2 \\ (\frac{3}{2}) \frac{2 x}{3} & = & \frac{2}{\log 8} (\frac{3}{2}) \end{array}$

$x = \frac{3}{\log 8}$

Note

You can start by simplifying $8^{\frac{2}{3} x} = {(\sqrt[3]{8})}^{2 x} = {(2)}^{2 x} = 4^{x}$

$9^{x} = 29$

$\begin{array}{rcl} \log 9^{x} & = & \log 29 \\ x \log 9 & = & \log 29 \end{array}$

$x = \frac{\log 29}{\log 9}$