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?
$\endgroup$ 35 Answers
$\begingroup$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.
$\endgroup$ $\begingroup$$\log_3 x=4\implies x=3^4=81,$ not $x=4$
$\endgroup$ $\begingroup$When you want to find the value of $~x~$ inside a logarithm function, first you have to make both side in term of logarithm first (convert four into a logarithm term) and then inverse it to find $~x~$ :
$\log_{3}(x) = 4(1)$ $\Leftarrow$ convert $~1~$ into logarithm base $~3~$
$\log_{3}(x) = 4.\log_{3}(3)$ $\Leftarrow$ raise $~4~$ as exponent
$\log_{3}(x) = \log_{3}(3^{4})$ $\Leftarrow$ inverse it using exponent function
$3^{\log_{3}(x)} = 3^{\log_{3}(3^{4})}$
$ x = 3^{4} = 81$
$\endgroup$ $\begingroup$First, $\log_{1/3}x = \log_{3}\frac{1}{x}$; hence,
$$ \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.$$
$\endgroup$ $\begingroup$$\log_3 (x)=\log_{1/3} (x)+8 $
$\iff \log_3 (x)=-\log_3 (x)+8$
$\iff 2\log_3 (x)=8$
$\iff \log_3 (x)=4$
$\implies x=3^4=81$
