Practice 1 Solutions

Convert each equation to the equivalent logarithmic form.

$27^{\frac{2}{3}} = 9$

$\log_{27} 9 = \frac{2}{3}$

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

$\log_{49} \frac{1}{7} = - \frac{1}{2}$

$15^{2} = 225$

$\log_{15} 225 = 2$

$9^{0} = 1$

$\log_{9} 1 = 0$

Convert each equation to the equivalent exponential form.

$\log_{16} 2 = \frac{1}{4}$

$16^{\frac{1}{4}} = 2$

$\log_{3} 3 = 1$

$3^{1} = 3$

$\log_{5} \sqrt[3]{5} = \frac{1}{3}$

$5^{\frac{1}{3}} = \sqrt[3]{5}$

$\log_{4} 64 = 3$

$4^{3} = 64$

Evaluate.

$\log_{4} 8$

$\log_{4} 8 = x 4^{x} = 8 2^{2 (x)} = 2^{3} 2 x = 3 x = \frac{3}{2}$

$\frac{3}{2}$

$\log_{27} 81$

$\log_{27} 81 = x 27^{x} = 81 3^{3 (x)} = 3^{4} 3 x = 4 x = \frac{4}{3}$

$\frac{4}{3}$

$\log_{5} 625$

$\log_{5} 625 = x 5^{x} = 625 5^{x} = 5^{4} x = 4$

$\log_{\frac{1}{5}} 25$

$\log_{\frac{1}{5}} 25 = x {(\frac{1}{5})}^{x} = 25 5^{- x} = 5^{2} - x = 2 x = - 2$

–2

$\log_{49} \frac{1}{343}$

$\log_{49} \frac{1}{343} = x 49^{x} = \frac{1}{343} 7^{2 (x)} = 7^{- 3} 2 x = - 3 x = - \frac{3}{2}$

$- \frac{3}{2}$

$\log_{4} \frac{1}{32}$

$\log_{4} \frac{1}{32} = x 4^{x} = \frac{1}{32} 2^{2 (x)} = 2^{- 5} 2 x = - 5 x = - \frac{5}{2}$

$- \frac{5}{2}$

$\log_{9} 27$

$\log_{9} 27 = x 9^{x} = 27 3^{2 (x)} = 3^{3} 2 x = 3 x = \frac{3}{2}$

$\frac{3}{2}$

$\log_{2} 64$

$\log_{2} 64 = x 2^{x} = 64 2^{x} = 2^{6} x = 6$

Evaluate with a calculator to the ten thousandths (four decimal places).

$\log 100$

2.0000

$\log 295$

2.4698

$\log 16$

1.2041

$\log 84$

1.9243