# 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?

• $\log_{3}(x) = -\log_{3}(x) + 8$ does not imply $0 = 8$. – Martin R Jul 15 '19 at 11:59
• After your edit: $\log_3(x) = 4$ does not imply $x=4$. – Martin R Jul 15 '19 at 12:05
• Do you remember the definition of the logarithmic function? $$\log_{a}{x}=b\Longleftrightarrow a^b=x.$$ – Michael Rybkin Jul 16 '19 at 10:31

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.

$$\log_3 x=4\implies x=3^4=81,$$ not $$x=4$$

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

$$3^{\log_{3}(x)} = 3^{\log_{3}(3^{4})}$$

$$x = 3^{4} = 81$$

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,

$$x = 3^{4} = 81.$$

$$\log_3 (x)=\log_{1/3} (x)+8$$
$$\iff \log_3 (x)=-\log_3 (x)+8$$
$$\iff 2\log_3 (x)=8$$
$$\iff \log_3 (x)=4$$
$$\implies x=3^4=81$$
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