Practice 2 Solutions

Expand.

$\log_{4} \frac{\sqrt[3]{b}}{4}$

$\log_{4} \frac{b^{\frac{1}{3}}}{4} \log_{4} b^{\frac{1}{3}} - \log_{4} 4$

$\frac{1}{3} \log_{4} b - 1$

Note

Recall, $\log_{a} (a^{x}) = x,$ therefore, $\log_{4} 4 = 1 .$

$\log_{3} \frac{10 x^{3} \sqrt{y}}{z^{5}}$

$\log_{3} \frac{10 x^{3} y^{\frac{1}{2}}}{z^{5}} \log_{3} 10 x^{3} y^{\frac{1}{2}} - \log_{3} z^{5} \log_{3} 10 + \log_{3} x^{3} + \log_{3} y^{\frac{1}{2}} - \log_{3} z^{5}$

$\log_{3} 10 + 3 \log_{3} x + \frac{1}{2} \log_{3} y - 5 \log_{3} z$

$\log \frac{x^{2} + 2 x + 1}{4 x^{2} - 9}$

$\log (\frac{(x + 1) (x + 1)}{(2 x - 3) (2 x + 3)}) \log (\frac{{(x + 1)}^{2}}{(2 x - 3) (2 x + 3)}) \log {(x + 1)}^{2} - \log (2 x - 3) (2 x + 3) 2 \log (x + 1) - [\log (2 x - 3) + \log (2 x + 3)]$

$2 \log (x + 1) - \log (2 x - 3) - \log (2 x + 3)$

Note

Factor each quadratic expression to expand completely.

$\log_{7} \frac{1}{5 x}$

$\log_{7} 1 - \log_{7} 5 x \log_{7} 1 - [\log_{7} 5 + \log_{7} x] \log_{7} 1 - \log_{7} 5 - \log_{7} x$

$- \log_{7} 5 - \log_{7} x$

Note

Recall, $\log_{a} 1 = 0,$ therefore, $\log_{7} 1 = 0$ .

Contract. Rewrite as a single log.

$\frac{2}{3} \log (x - 4) - \log (x + 1)$

$\log {(x - 4)}^{\frac{2}{3}} - \log (x + 1) \log \frac{{(x - 4)}^{\frac{2}{3}}}{x + 1} \log \frac{{({(x - 4)}^{2})}^{\frac{1}{3}}}{x + 1}$

$\log \frac{\sqrt[3]{{(x - 4)}^{2}}}{x + 1}$

$5 \log_{8} x + 3 \log_{8} y - 2 \log_{8} z - 6 \log_{8} a$

$\log_{8} x^{5} + \log_{8} y^{3} - \log_{8} z^{2} - \log_{8} a^{6} \log_{8} x^{5} y^{3} - (\log_{8} z^{2} + \log_{8} a^{6}) \log_{8} x^{5} y^{3} - \log_{8} z^{2} a^{6}$

$\log_{8} \frac{x^{5} y^{3}}{z^{2} a^{6}}$

$\frac{2}{3} \log_{2} x + \frac{1}{3} \log_{2} y + \frac{2}{3} \log_{2} z$

$\log_{2} x^{\frac{2}{3}} + \log_{2} y^{\frac{1}{3}} + \log_{2} z^{\frac{2}{3}} \log_{2} x^{\frac{2}{3}} y^{\frac{1}{3}} z^{\frac{2}{3}} \log_{2} {(x^{2} y^{1} z^{2})}^{\frac{1}{3}}$

$\log_{2} \sqrt[3]{x^{2} y z^{2}}$

Note

The final answer should be in radical form.

$3 \log (x - 5) + 2 \log (x + 2) - 4 \log 2 x$

$\log {(x - 5)}^{3} + \log {(x + 2)}^{2} - \log {(2 x)}^{4} \log {(x - 5)}^{3} {(x + 2)}^{2} - \log {(2 x)}^{4}$

$\log \frac{{(x - 5)}^{3} {(x + 2)}^{2}}{{(2 x)}^{4}}$

Solve.

$\log_{3} (4 x + 5) - \log_{3} x = 2$

$\begin{array}{rcl} \log_{3} \frac{4 x + 5}{x} & = & 2 \\ 3^{2} & = & \frac{4 x + 5}{x} \\ 9 x & = & 4 x + 5 \\ 5 x & = & 5 \end{array}$

x = 1

$\log_{4} x + \log_{4} (2 x - 3) = \log_{4} 5$

$\begin{array}{rcl} \log_{4} [x (2 x - 3)] & = & \log_{4} 5 \\ x (2 x - 3) & = & 5 \\ 2 x^{2} - 3 x & = & 5 \\ 2 x^{2} - 3 x - 5 & = & 0 \\ (x + 1) (2 x - 5) & = & 0 \\ x & = & - 1, \frac{5}{2} \end{array}$

$x = \frac{5}{2}$

$\log_{9} (3 x - 7) = \log_{9} x + \log_{9} 2$

$\begin{array}{rcl} \log_{9} (3 x - 7) & = & \log_{9} (2 x) \\ 3 x - 7 & = & 2 x \end{array}$

x = 7

$\log_{36} (4 x - 3) = \frac{1}{2}$

$\begin{array}{rcl} 36^{\frac{1}{2}} & = & 4 x - 3 \\ 6 & = & 4 x - 3 \\ 4 x & = & 9 \end{array}$

$x = \frac{9}{4}$

$\log_{2} (x - 3) + \log_{2} (x - 5) = \log_{2} 3$

$\log_{2} (x - 3) (x - 5) = \log_{2} 3 (x - 3) (x - 5) = 3 x^{2} - 8 x + 15 = 3 x^{2} - 8 x + 12 = 0 (x - 2) (x - 6) = 0 x = 2, 6$

x = 6

$\log_{5} (x - 1) + \log_{5} (x + 3) - 2 \log_{5} x = 0$

$\begin{array}{rcl} \log_{5} [(x - 1) (x + 3)] - \log_{5} x^{2} & = & 0 \\ \log_{5} \frac{(x - 1) (x + 3)}{x^{2}} & = & 0 \\ 5^{0} & = & \frac{(x - 1) (x + 3)}{x^{2}} \\ 1 & = & \frac{(x - 1) (x + 3)}{x^{2}} \\ x^{2} & = & x^{2} + 2 x - 3 \\ 0 & = & 2 x - 3 \\ 2 x & = & 3 \end{array}$

$x = \frac{3}{2}$

Note

This equation can also be solved by rewriting the equation: $\log_{5} (x - 1) + \log_{5} (x + 3) = 2 \log_{5} x$

$\log_{4} (3 x + 7) = 3$

$4^{3} = 3 x + 7 64 = 3 x + 7 57 = 3 x$

x = 19

$\log_{5} (x + 4) - \log_{5} (x - 2) = 2$

$\begin{array}{rcl} \log_{5} \frac{x + 4}{x - 2} & = & 2 \\ 5^{2} & = & \frac{x + 4}{x - 2} \\ 25 (x - 2) & = & x + 4 \\ 25 x - 50 & = & x + 4 \\ 24 x & = & 54 \end{array}$

$x = \frac{9}{4}$

$\log (x - 5) - \log (x + 4) = \log (x - 1) - \log (x + 2)$

$\begin{array}{rcl} \log \frac{x - 5}{x + 4} & = & \log \frac{x - 1}{x + 2} \\ \frac{x - 5}{x + 4} & = & \frac{x - 1}{x + 2} \\ (x - 5) (x + 2) & = & (x - 1) (x + 4) \\ x^{2} - 3 x - 10 & = & x^{2} + 3 x - 4 \\ - 6 x & = & 6 \\ x & = & - 1 \end{array}$

no solution

Note

When $x = - 1$ , the arguments $(x - 5)$ and $(x - 1)$ are not greater than zero.

$2 \log_{4} (x - 2) = \log_{4} x + \log_{4} (x - 3)$

$\begin{array}{rcl} \log_{4} {(x - 2)}^{2} & = & \log_{4} (x (x - 3)) \\ {(x - 2)}^{2} & = & x (x - 3) \\ x^{2} - 4 x + 4 & = & x^{2} - 3 x \\ - 4 x + 4 & = & - 3 x \end{array}$

x = 4

$\frac{1}{2} \log_{2} (x + 3) = 2$

$\begin{array}{rcl} (2) (\frac{1}{2} \log_{2} (x + 3)) & = & (2) (2) \\ \log_{2} (x + 3) & = & 4 \\ 2^{4} & = & x + 3 \\ 16 & = & x + 3 \end{array}$

x = 13

Note

An alternate way of solving:
$\begin{array}{rcl} \log_{2} {(x + 3)}^{\frac{1}{2}} & = & 2 \\ 2^{2} & = & {(x + 3)}^{\frac{1}{2}} \\ {(4)}^{2} & = & {(\sqrt{x + 3})}^{2} \\ 16 & = & x + 3 \end{array}$

$- 5 \log_{3} (2 x + 5) = - 15$

$\begin{array}{rcl} \log_{3} (2 x + 5) & = & 3 \\ 3^{3} & = & 2 x + 5 \\ 27 & = & 2 x + 5 \\ 2 x & = & 22 \end{array}$

x = 11