Question

How do I evaluate a log when the argument is in a fraction?

Answer 1

Explanation:

Given that

$\begin{gathered} \log_43\approx0.7925 \\ find \\ \log_4(\frac{1}{9}) \end{gathered}$

Apply the law of fractional indices below

$\frac{1}{a}=a^{-1}$

By applying the law, we will have that

$\begin{gathered} \frac{1}{9}=9^{-1} \\ recall: \\ 9=3^2 \\ 9^{-1}=3^{-2} \end{gathered}$

By rewriting the expression, we will have

$\operatorname{\log}_4(\frac{1}{9})=\log_4(3^{-2})$

Apply the logarithmi law of exponents below

$\log_ba^c=c\log_ba$$\begin{gathered} \log_4(3^{-2})=-2\log_43 \\ -2\log_43=-2(0.7925) \\ =-1.5850 \end{gathered}$

Hence,

The final answer to 4 decimal places is

$-1.5850$