Solve 9^x-4=81

$$9 ^ { x - 4 } = 81$$

Solve for x

$$x=6$$

Steps Using Definition of Logarithm

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

$$9^{x-4}=81$$

Take the logarithm of both sides of the equation.

$$\log(9^{x-4})=\log(81)$$

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

$$\left(x-4\right)\log(9)=\log(81)$$

Divide both sides by \log(9).

$$x-4=\frac{\log(81)}{\log(9)}$$

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

$$x-4=\log_{9}\left(81\right)$$

Add 4 to both sides of the equation.

$$x=2-\left(-4\right)$$

Solve for x (complex solution)

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

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