Solve 9^x-1=27

$$9 ^ { x - 1 } = 27$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

Steps Using Definition of Logarithm

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

$$9^{x-1}=27$$

Take the logarithm of both sides of the equation.

$$\log(9^{x-1})=\log(27)$$

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

$$\left(x-1\right)\log(9)=\log(27)$$

Divide both sides by \log(9).

$$x-1=\frac{\log(27)}{\log(9)}$$

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

$$x-1=\log_{9}\left(27\right)$$

Add 1 to both sides of the equation.

$$x=\frac{3}{2}-\left(-1\right)$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(3)}+\frac{5}{2}$$

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