# Math Help - Log

1. ## Log

2. since $\log_{3}(3)=1$, $\log_{9}(9)=1$ and $1=\log_{4}(4)$, the equation can be written like the following:
$2\log_{4}(3x)-\log_{4}(x)=\log_{4}(4)-\frac{1}{2}\log_{4}(x^2)$
since $a\log(b)=\log(b)^a$, then:
$\log_{4}(3x)^2-\log_{4}(x)=\log_{4}(4)-\log_{4}(x^2)^{\frac{1}{2}}$
since $(x^2)^{\frac{1}{2}}=x$ and $\log(a)-\log(b)=\log(\frac{a}{b})$
then the equation can be written like:
$\log_{4}(\frac{(3x)^2}{x})=\log_{4}(\frac{4}{x})$
$\log_{4}(9x)=\log_{4}(\frac{4}{x})$

Now, Solve the last equation.

3. Hello, reiward!

$2\log_4(3x)\log_3(3) - \log_4(x) \;=\;1 - \tfrac{1}{2}\log_4(x^2)\log_9(9)$

I have it down to this and I'm lost: . $2\log_4(3x) \:=\:1$ . .
. . . Keep going!

$\text{You have: }\;\;\log_4(3x)^2 \;=\;1$

. . . . . . . . $\log_4(9x^2) \;=\;1$

. . . . . . . . . . . $9x^2 \;=\;4^1$

. . . . . . . . . . . $x^2 \;=\;\frac{4}{9}$

. . . . . . . . . . . . $x \;=\;\frac{2}{3}$

(The negative square root is an extraneous root.)

4. Originally Posted by reiward
$2\log_4(3x)=1\Longrightarrow\log_4(3x)^2=1\Longrig htarrow (3x)^2=4^1$ ...take it from here