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?
Best Answer
The mistake:
It should be $$2\log_3x=8$$ or $$\log_3x=4$$ or $$\log_3x=\log_3{3^4},$$ which gives $x=81$.
Actually, in the first step you can use the following property. $$\log_{a^{\beta}}x=\frac{1}{\beta}\log_ax,$$ where $a>0$, $a\neq1$, $x>0$ and $\beta\neq0$.
Since $\frac{1}{3}=3^{-1},$ we obtain $$\log_3x=-\log_3x+8$$ immediately.