<s>[INST] How to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$ I'm trying to solve $\log_{3}(x) = \log_{\frac{1}{3}}(x) + 8$.  I am getting x = 4 but the book gets x = 81.  What am I doing wrong? 
\begin{align*}
\log_{3}(x) & = \log_{\frac{1}{3}}(x) + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{\log_{3}(\frac{1}{3})} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-\log_{3}(3)} + 8\\
\log_{3}(x) & = \frac{\log_{3}(x)}{-1} + 8\\
\log_{3}(x) & = -\log_{3}(x) + 8\\
2\log_{3}(x) & =  8\\
\log_{3}(x) & =  4\\
x & =  4
\end{align*}
What am I doing wrong?
 [/INST] First, $\log_{1/3}x = \log_{3}\frac{1}{x}$; hence,
$$ \log_{3}x - \log_{3} \left( \frac{1}{x} \right)  = \log_{3} \left(\frac{x}{\frac{1}{x}} \right) = \log_{3}x^{2} = 8$$
which is equivalent to
$$ 3^{8} = \left( 3^{4} \right)^{2} = x^{2}.$$
And so, 
$$x = 3^{4} = 81.$$
</s>