If:
$\displaystyle 6^{2x} = 3^{x+1}$
Prove that $\displaystyle x = \;\log_{12}3$
Any tips on identity's and proving would be helpful thanks i seem to struggle with them.
If:
$\displaystyle 6^{2x} = 3^{x+1}$
Prove that $\displaystyle x = \;\log_{12}3$
Any tips on identity's and proving would be helpful thanks i seem to struggle with them.
$\displaystyle 6^{2x}=3^{x+1}\Rightarrow 36^x=3\cdot 3^x\Rightarrow\log_{12}36^x=\log_{12}(3\cdot 3^x)\Rightarrow$
$\displaystyle \Rightarrow x\log_{12}36=\log_{12}3+x\log_{12}3\Rightarrow x\left(\log_{12}36-\log_{12}3\right)=\log_{12}3\Rightarrow$
$\displaystyle \Rightarrow x\cdot\log_{12}\frac{36}{3}=\log_{12}3\Rightarrow x\log_{12}12=\log_{12}3\Rightarrow x=\log_{12}3$