How do I turn $\displaystyle 3^x$ = $\displaystyle 4^x^-^2$ into $\displaystyle \frac{\log\frac1{16}}{log\frac3{4}}$ and could you please also explain the steps thoroughly. Thanks!
1st, you don't. The first is an EQUATION and the second is an EXPRESSION. Not the same at all.
2nd, this is a logarithm problem, as you can see by the final form. If everything is suitable, then $\displaystyle a^{b} = c^{d}$ implies $\displaystyle b\cdot log(a) = d\cdot log(c)$. That's almost all you need.
Hello, Jayissocool!
First of all, that is a really stupid way to leave the answer!
$\displaystyle \text{How do I turn }\,3^x \:=\:4^x^{-2}\,\text{ into }\,x \:=\:\frac{\log\frac1{16}}{\log\frac3{4}}\;?$
Take logs: .$\displaystyle \log\left(3^x) \;=\;\log\left(4^{x-2}\right) \quad\Rightarrow\quad x\log3 \;=\;(x-2)\log4$
. . $\displaystyle x\log 3 \;=\;x\log4 - 2\log 4 \quad\Rightarrow\quad x\log 3 - x\log4 \;=\;-2\log 4$
. . $\displaystyle x(\log3-\log4) \;=\;\log(4^{-2}) \quad\Rightarrow\quad x \;=\;\frac{\log(4^{-2})}{\log3-\log 4}$
Therefore: .$\displaystyle x \;=\;\dfrac{\log\frac{1}{16}}{\log\frac{3}{4}} \;\;\hdots\text{ which equals: }\,x \,=\,\frac{\log16}{\log\frac{4}{3}}$