## Precalculus: Mathematics for Calculus, 7th Edition

$\log_{49} (\frac{1}{7}) = -\frac{1}{2}$
$Write$ $the$ $equation$ $in$ $logarithmic$ $form:$ $49^{-\frac{1}{2}} = \frac{1}{7}$ Exponential form to logarithmic form: $b^c = a \rightarrow \log_b a = c$ Plug in 49 for b, $-\frac{1}{2}$ for c, and $\frac{1}{7}$ for a $$49^{-\frac{1}{2}} = \frac{1}{7} \rightarrow \log_{49} (\frac{1}{7}) = -\frac{1}{2}$$