Answer
The two problems are equivalent - they are equally "easy" or "hard".
In either case, we are seeking the exponent x by which 9 must be raised in order to obtain $\displaystyle \frac{1}{3}.$
This is because we use the definition of logarithm to rewrite a logarithmic equation as an equivalent exponential equation or the other way around:
$m=\log_{a}x$ is equivalent to $a^{m}=x.$
Here, because $3^{2}=9\displaystyle \ \Rightarrow\ 3=9^{1/2}\ \Rightarrow\ \frac{1}{3}=9^{-1/2}$, we have
$x=-\displaystyle \frac{1}{2}$,
as $9^{-1/2}=\displaystyle \frac{1}{3}$, or, equivalently, $\displaystyle \log_{9}(9^{-1/2})=-\frac{1}{2}$.
Work Step by Step
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