Solve 1/3=9/8*(2/3)^n-1

$$\frac { 1 } { 3 } = \frac { 9 } { 8 } \times ( \frac { 2 } { 3 } ) ^ { n - 1 }$$

Solve for n

$$n=4$$

Steps Using Definition of Logarithm

Multiply both sides by \frac{8}{9}, the reciprocal of \frac{9}{8}.

$$\frac{1}{3}\times \frac{8}{9}=\left(\frac{2}{3}\right)^{n-1}$$

Multiply \frac{1}{3} and \frac{8}{9} to get \frac{8}{27}.

$$\frac{8}{27}=\left(\frac{2}{3}\right)^{n-1}$$

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

$$\left(\frac{2}{3}\right)^{n-1}=\frac{8}{27}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{2}{3}\right)^{n-1})=\log(\frac{8}{27})$$

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

$$\left(n-1\right)\log(\frac{2}{3})=\log(\frac{8}{27})$$

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

$$n-1=\frac{\log(\frac{8}{27})}{\log(\frac{2}{3})}$$

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

$$n-1=\log_{\frac{2}{3}}\left(\frac{8}{27}\right)$$

Add 1 to both sides of the equation.

$$n=3-\left(-1\right)$$