Solve 3^2x-3=9

$$3 ^ { 2 x - 3 } = 9$$

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.

$$3^{2x-3}=9$$

Take the logarithm of both sides of the equation.

$$\log(3^{2x-3})=\log(9)$$

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

$$\left(2x-3\right)\log(3)=\log(9)$$

Divide both sides by \log(3).

$$2x-3=\frac{\log(9)}{\log(3)}$$

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

$$2x-3=\log_{3}\left(9\right)$$

Add 3 to both sides of the equation.

$$2x=2-\left(-3\right)$$

Divide both sides by 2.

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

Solve for x (complex solution)

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

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