1. ## Log problem again

Hi all, I have this log problem that I having trouble getting $\log_{8} \frac {x}{2} = \frac {\log_{8}x}{\log_{8}2}$ can someone please help me solve this log equation.

2. By changing the base of log..

$\frac{\log_{e}\frac{x}{2}}{\log_{e}{8}}=\frac{\log _{e}x}{\log_{e}2}$

$\therefore\frac{\log_{e}\frac{x}{2}}{3\log_{e}{2}} =\frac{\log_{e}x}{\log_{e}2}$

$\therefore\log_{e}x-\log_{e}2=3\log_{e}x$

$\therefore\log_{e}x-\log_{e}2=\log_{e}x^3$

$\therefore\log_{e}x-\log_{e}x^3=\log_{e}2$

$\therefore\log_{e}\frac{x}{x^3}=\log_{e}2$

$\therefore x=2x^3$

$\therefore 2x^3-x=0$

Now solve this Equation....

3. Hello, scrible!

Solve for $x\!:\;\;\log_8\left(\frac {x}{2}\right) \;=\; \frac{\log_8(x)}{\log_8(2)}$
First, evaluate $\log_8(2)$

Let: . $\log_8(2) \:=\:P \quad\Rightarrow\quad 8^P \:=\:2 \quad\Rightarrow\quad \left(2^3\right)^P \:=\:2 \quad\Rightarrow \quad 2^{3P} \:=\:2^1$

Hence: . $3P \:=\:1 \quad\Rightarrow\quad P \:=\:\tfrac{1}{3}$

Therefore: . $\log_8(2) \,=\,\tfrac{1}{3}$ .[1]

The equation is: . $\log_8(x) - \log_8(2) \;=\;\frac{\log_8(x)}{\log_8(2)}$

Substitute [1]: . $\log_8(x) - \tfrac{1}{3} \;=\;\frac{\log_8(x)}{\frac{1}{3}} \quad\Rightarrow\quad \log_8(x) - \tfrac{1}{3} \;=\;3\log_8(x)$

Multiply by 3: . $3\log_8(x) - 1 \;=\;9\log_8(x) \quad\Rightarrow\quad -6\log_8(x) \;=\;1 \quad\Rightarrow\quad \log_8(x) \;=\;-\tfrac{1}{6}$

Hence: . $x \;=\;8^{-\frac{1}{6}} \;=\;\left(2^3\right)^{-\frac{1}{6}} \;=\;2^{-\frac{1}{2}}$

Therefore: . $\boxed{x \;=\;\frac{1}{\sqrt{2}}}$

4. Originally Posted by Soroban
Hello, scrible!

First, evaluate $\log_8(2)$

Let: . $\log_8(2) \:=\:P \quad\Rightarrow\quad 8^P \:=\:2 \quad\Rightarrow\quad \left(2^3\right)^P \:=\:2 \quad\Rightarrow \quad 2^{3P} \:=\:2^1$

Hence: . $3P \:=\:1 \quad\Rightarrow\quad P \:=\:\tfrac{1}{3}$

Therefore: . $\log_8(2) \,=\,\tfrac{1}{3}$ .[1]

The equation is: . $\log_8(x) - \log_8(2) \;=\;\frac{\log_8(x)}{\log_8(2)}$

Substitute [1]: . $\log_8(x) - \tfrac{1}{3} \;=\;\frac{\log_8(x)}{\frac{1}{3}} \quad\Rightarrow\quad \log_8(x) - \tfrac{1}{3} \;=\;3\log_8(x)$

Multiply by 3: . $3\log_8(x) - 1 \;=\;9\log_8(x) \quad\Rightarrow\quad -6\log_8(x) \;=\;1 \quad\Rightarrow\quad \log_8(x) \;=\;-\tfrac{1}{6}$

Hence: . $x \;=\;8^{-\frac{1}{6}} \;=\;\left(2^3\right)^{-\frac{1}{6}} \;=\;2^{-\frac{1}{2}}$

Therefore: . $\boxed{x \;=\;\frac{1}{\sqrt{2}}}$
Thank you, but when I checked the answer in the book they has the answer as being $\frac{\sqrt{2}}{2}$

5. Originally Posted by scrible
Thank you, but when I checked the answer in the book they has the answer as being $\frac{\sqrt{2}}{2}$
Formally, you should never have a square root in the denominator. Here's what the book did (multiply by 1) to get rid of the square root in the bottom.

$x = \frac{1}{\sqrt{2}} * \frac{\sqrt{2}}{\sqrt{2}}
= \frac{\sqrt{2}}{2}$