<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] When you want to find the value of $~x~$ inside a logarithm function, first you have to make both side in term of logarithm first (convert four into a logarithm term) and then inverse it to find $~x~$ :
$\log_{3}(x) = 4(1)$ $\Leftarrow$ convert $~1~$ into logarithm base $~3~$
$\log_{3}(x) = 4.\log_{3}(3)$ $\Leftarrow$ raise $~4~$ as exponent
$\log_{3}(x) = \log_{3}(3^{4})$ $\Leftarrow$ inverse it using exponent function
$3^{\log_{3}(x)} = 3^{\log_{3}(3^{4})}$
$ x = 3^{4} = 81$
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