# Thread: [SOLVED] Log proof

1. ## [SOLVED] Log proof

If:

$6^{2x} = 3^{x+1}$

Prove that $x = \;\log_{12}3$

Any tips on identity's and proving would be helpful thanks i seem to struggle with them.

2. This is not true.
For x=1 we have 36=9!!!

Maybe you have to solve the equation.

3. My apologies red dog I left out the rest of the question but I have fixed it up now

4. $6^{2x}=3^{x+1}\Rightarrow 36^x=3\cdot 3^x\Rightarrow\log_{12}36^x=\log_{12}(3\cdot 3^x)\Rightarrow$

$\Rightarrow x\log_{12}36=\log_{12}3+x\log_{12}3\Rightarrow x\left(\log_{12}36-\log_{12}3\right)=\log_{12}3\Rightarrow$

$\Rightarrow x\cdot\log_{12}\frac{36}{3}=\log_{12}3\Rightarrow x\log_{12}12=\log_{12}3\Rightarrow x=\log_{12}3$

5. Another way:
$\begin{array}{rcl}
{6^{2x} } & = & {2^{2x} \cdot 3^{2x} } \\
{2^{2x} \cdot 3^{2x} } & = & {3^{x + 1} } \\
{2^{2x} \cdot 3^x } & = & 3 \\
{12^x } & = & 3 \\
x & = & {\log _{12} (3)} \\

\end{array}$