Solve (2/3)^{n+1}*(2/3)^{2}=frac{6^{2}*128*81^{6}*2015^0}{(3^4)^5*8*27^4}

$$( \frac { 2 } { 3 } ) ^ { n + 1 } \cdot ( \frac { 2 } { 3 } ) ^ { 2 } = \frac { 6 ^ { 2 } \cdot 128 \cdot 81 ^ { 6 } \cdot 2015 ^ { 0 } } { ( 3 ^ { 4 } ) ^ { 5 } \cdot 8 \cdot 27 ^ { 4 } }$$

Solve for n

$$n=3$$

Steps Using Definition of Logarithm

Use the rules of exponents and logarithms to solve the equation.

$$\frac{4}{9}\times \left(\frac{2}{3}\right)^{n+1}=\frac{64}{729}$$

Divide both sides of the equation by \frac{4}{9}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\left(\frac{2}{3}\right)^{n+1}=\frac{16}{81}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{2}{3}\right)^{n+1})=\log(\frac{16}{81})$$

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

$$\left(n+1\right)\log(\frac{2}{3})=\log(\frac{16}{81})$$

Divide both sides by \log(\frac{2}{3}).

$$n+1=\frac{\log(\frac{16}{81})}{\log(\frac{2}{3})}$$

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

$$n+1=\log_{\frac{2}{3}}\left(\frac{16}{81}\right)$$

Subtract 1 from both sides of the equation.

$$n=4-1$$