Answer
$-2$
Work Step by Step
Recall
(1) $\log_a{b}=y \longleftrightarrow a^y=b$
(2) The value of the function $f(x)=\log_a{x}$ increases as $x$ increases.
Let $y=\log_{5}{\frac{1}{47}}$
Use the definition in (1) above to obtain:
$y=\log_{5}{\frac{1}{47}}\longrightarrow 5^y=\frac{1}{47}$
Note that
$5^{-2}=\frac{1}{5^2}=\frac{1}{25}$
$5^{-3}=\frac{1}{5^3} = \frac{1}{125}$
Since $\frac{1}{125} \lt \frac{1}{47} \lt \frac{1}{25}$, then $y$ must be between $-3$ and $-2$.
Thus, the least integer that is greater than $\log_{5}{\frac{1}{47}}$ is $-2$.
