Practice 2 Solutions

Convert the equation to its inverse (exponential to logarithmic or logarithmic to exponential).

$5^{- 2} = \frac{1}{25}$

$\log_{5} \frac{1}{25} = - 2$

$64^{\frac{2}{3}} = 16$

$\log_{64} 16 = \frac{2}{3}$

$49^{- \frac{3}{2}} = \frac{1}{343}$

$\log_{49} \frac{1}{343} = - \frac{3}{2}$

$6^{- 2} = \frac{1}{36}$

$\log_{6} \frac{1}{36} = - 2$

$\log_{23} 1 = 0$

$23^{0} = 1$

$\log_{17} \sqrt{17} = \frac{1}{2}$

$17^{\frac{1}{2}} = \sqrt{17}$

$\log_{196} 14 = \frac{1}{2}$

$196^{\frac{1}{2}} = 14$

$\log_{20} \frac{1}{400} = - 2$

$20^{- 2} = \frac{1}{400}$

Evaluate.

$\log_{11} \sqrt[4]{11}$

$\log_{11} \sqrt[4]{11} = x 11^{x} = \sqrt[4]{11} 11^{x} = 11^{\frac{1}{4}} x = \frac{1}{4}$

$\frac{1}{4}$

$\log_{6} 216$

$\log_{6} 216 = x 6^{x} = 216 6^{x} = 6^{3} x = 3$

$\log_{225} 15$

$\log_{225} 15 = x 225^{x} = 15 15^{2 (x)} = 15^{1} 2 x = 1 x = \frac{1}{2}$

$\frac{1}{2}$

$\log_{12} \frac{1}{144}$

$\log_{12} \frac{1}{144} = x 12^{x} = \frac{1}{144} 12^{x} = 12^{- 2} x = - 2$

–2

$\log_{25} 125$

$\log_{25} 125 = x 25^{x} = 125 5^{2 (x)} = 5^{3} 2 x = 3 x = \frac{3}{2}$

$\frac{3}{2}$

$\log_{32} 64$

$\log_{32} 64 = x 32^{x} = 64 2^{5 (x)} = 2^{6} 5 x = 6 x = \frac{6}{5}$

$\frac{6}{5}$

$\log_{81} 243$

$\log_{81} 243 = x 81^{x} = 243 3^{4 (x)} = 3^{5} 4 x = 5 x = \frac{5}{4}$

$\frac{5}{4}$

$\log_{0.25} 4$

$\log_{0.25} 4 = x {(\frac{1}{4})}^{x} = 4 4^{- 1 (x)} = 4^{1} - x = 1 x = - 1$

–1

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

$\log 7$

0.8451

$\log 1$

Note

Recall: $\log_{b} 1 = 0$ because $b^{0} = 1$

$\log (0.001)$

–2

$\log 1225$

3.0881