How to approach solving this indices example? I am having problems working out how to approach solving this problem :
$$\left(\frac{81}{16}\right)^n = \frac{32}{243}$$
How do I go about working out $^n$? Please step by step if possible.
 A: Using laws of exponents, you can rewrite the equation as follows.
$$\left(\frac{81}{16}\right)^n = \frac{32}{243}$$
$$\left(\frac{3^4}{2^4}\right)^n = \frac{2^5}{3^5}$$
$$\left(\frac{3}{2}\right)^{4n} = \left(\frac{2}{3}\right)^5 $$
$$\left(\frac{2}{3}\right)^{-4n} = \left(\frac{2}{3}\right)^5$$
From this, it follows that
$$-4n = 5$$
$$n = -\frac{5}{4}$$
A: Here is an approach.
$\displaystyle \left(\frac{81}{16}\right)^n = \frac{32}{243}$
$\displaystyle \left(\frac{3^4}{2^4}\right)^n = \left(\frac{2^5}{3^5}\right)$
$\displaystyle \left(\frac{2}{3}\right)^{-4n} = \left(\frac{2}{3}\right)^5$
So, $-4n = 5 \rightarrow n = -\frac{5}{4}$
A: apply $\log$ on both side and reduce the equation
$$n \log\left(\frac{81}{16}\right) = \log\left(\frac{32}{243}\right) $$
$$ n(\log 81 -\log 16) = \log 32 - \log 243$$
$$ n = \frac{\log 32 - \log 243}{\log 81 -\log 16} $$
A: The point is $3^4=81$, $2^4=16$, $3^5=243$, and $2^5=32$:
$$\left(\frac{81}{16}\right)^n = \frac{32}{243}$$
$$
\left(\frac 3 2\right)^{4n} = \left(\frac 3 2 \right)^{-5}
$$
So $4n=-5$.
