Test 20 (Lessons 39–40): Logarithmic Expressions and Properties of Logs Solutions

Write $\log_{49} 7 = \frac{1}{2}$ in exponential form.

$49^{\frac{1}{2}} = 7$

Write $32^{\frac{3}{5}} = 8$ in logarithmic form.

$\log_{32} 8 = \frac{3}{5}$

Evaluate. Show your work.

$\log_{6} 216$

$\begin{array}{rcl} \log_{6} 216 & = & x \\ 6^{x} & = & 216 \\ 6^{x} & = & 6^{3} \\ x & = & 3 \end{array}$

$\log_{8} (\frac{1}{4})$

$\begin{array}{rcl} \log_{8} (\frac{1}{4}) & = & x \\ 8^{x} & = & \frac{1}{4} \\ 2^{3 (x)} & = & 2^{- 2} \\ 3 x & = & - 2 \\ x & = & - \frac{2}{3} \end{array}$

Expand: $\log \frac{2 \sqrt{x}}{5 x + 1}$

$\log \frac{2 {(x)}^{\frac{1}{2}}}{5 x + 1} \log 2 {(x)}^{\frac{1}{2}} - \log (5 x + 1) \log 2 + \log {(x)}^{\frac{1}{2}} - \log (5 x + 1) \log 2 + \frac{1}{2} \log x - \log (5 x + 1)$

Write $\frac{1}{4} \cdot 5 \log_{a} x + \frac{1}{4} \cdot 3 \log_{a} y - 8 \log_{a} z$ as a single logarithm.

$\frac{1}{4} \cdot 5 \log_{a} x + \frac{1}{4} \cdot 3 \log_{a} y - 8 \log_{a} z \frac{1}{4} (5 \log_{a} x + 3 \log_{a} y) - 8 \log_{a} z \frac{1}{4} (\log_{a} x^{5} + \log_{a} y^{3}) - \log_{a} z^{8} {(\log_{a} x^{5} y^{3})}^{\frac{1}{4}} - \log_{a} z^{8} \log_{a} \frac{{(x^{5} y^{3})}^{\frac{1}{4}}}{z^{8}} or \log_{a} \frac{\sqrt[4]{x^{5} y^{3}}}{z^{8}}$

Solve.

$3 \log_{8} 2 x = 2$

$\begin{array}{rcl} \log_{8} {(2 x)}^{3} & = & 2 \\ 8^{2} & = & 8 x^{3} \\ 64 & = & 8 x^{3} \\ 8 & = & x^{3} \\ x & = & 2 \end{array}$

$\log_{8} 6 = \log_{8} (x + 5) + \log_{8} x$

$\begin{array}{rcl} \log_{8} 6 & = & \log_{8} (x (x + 5)) \\ 6 & = & x (x + 5) \\ 0 & = & x^{2} + 5 x - 6 \\ 0 & = & (x - 1) (x + 6) \\ x & = & - 6, 1 \end{array}$

$\log_{4} (x + 4) - \log_{4} (x - 2) = 1$

$\begin{array}{rcl} \log_{4} (\frac{x + 4}{x - 2}) & = & 1 \\ 4^{1} & = & \frac{x + 4}{x - 2} \\ 4 (x - 2) & = & x + 4 \\ 4 x - 8 & = & x + 4 \\ 3 x & = & 12 \\ x & = & 4 \end{array}$

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

$\begin{array}{rcl} \log (3 x - 5) & = & \log 2 x \\ 3 x - 5 & = & 2 x \\ - 5 & = & - x \\ x & = & 5 \end{array}$