Solve (3/4)^-6*(4/3)^-8=(3/4)^n

$$( \frac { 3 } { 4 } ) ^ { - 6 } \times ( \frac { 4 } { 3 } ) ^ { - 8 } = ( \frac { 3 } { 4 } ) ^ { n }$$

Solve for n

$$n=2$$

Steps Using Definition of Logarithm

Calculate \frac{3}{4} to the power of -6 and get \frac{4096}{729}.

$$\frac{4096}{729}\times \left(\frac{4}{3}\right)^{-8}=\left(\frac{3}{4}\right)^{n}$$

Calculate \frac{4}{3} to the power of -8 and get \frac{6561}{65536}.

$$\frac{4096}{729}\times \frac{6561}{65536}=\left(\frac{3}{4}\right)^{n}$$

Multiply \frac{4096}{729} and \frac{6561}{65536} to get \frac{9}{16}.

$$\frac{9}{16}=\left(\frac{3}{4}\right)^{n}$$

Swap sides so that all variable terms are on the left hand side.

$$\left(\frac{3}{4}\right)^{n}=\frac{9}{16}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{3}{4}\right)^{n})=\log(\frac{9}{16})$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$n\log(\frac{3}{4})=\log(\frac{9}{16})$$

Divide both sides by \log(\frac{3}{4}).

$$n=\frac{\log(\frac{9}{16})}{\log(\frac{3}{4})}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$n=\log_{\frac{3}{4}}\left(\frac{9}{16}\right)$$