Solve 2(3^n-1/3-1)=728

$$2 ( \frac { 3 ^ { n } - 1 } { 3 - 1 } ) = 728$$

Solve for n

$$n=6$$

Steps Using Definition of Logarithm

Divide both sides by 2.

$$\frac{3^{n}-1}{3-1}=\frac{728}{2}$$

Divide 728 by 2 to get 364.

$$\frac{3^{n}-1}{3-1}=364$$

Subtract 1 from 3 to get 2.

$$\frac{3^{n}-1}{2}=364$$

Divide each term of 3^{n}-1 by 2 to get \frac{1}{2}\times 3^{n}-\frac{1}{2}.

$$\frac{1}{2}\times 3^{n}-\frac{1}{2}=364$$

Add \frac{1}{2} to both sides of the equation.

$$\frac{1}{2}\times 3^{n}=\frac{729}{2}$$

Multiply both sides by 2.

$$3^{n}=729$$

Take the logarithm of both sides of the equation.

$$\log(3^{n})=\log(729)$$

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

$$n\log(3)=\log(729)$$

Divide both sides by \log(3).

$$n=\frac{\log(729)}{\log(3)}$$

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

$$n=\log_{3}\left(729\right)$$

Solve for n (complex solution)

$$n=\frac{2\pi n_{1}i}{\ln(3)}+6$$

$$n_{1}\in \mathrm{Z}$$