Solve 9^3-x=81

$$9 ^ { 3 - x } = 81$$

Solve for x

$$x=1$$

Steps Using Definition of Logarithm

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

$$9^{-x+3}=81$$

Take the logarithm of both sides of the equation.

$$\log(9^{-x+3})=\log(81)$$

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

$$\left(-x+3\right)\log(9)=\log(81)$$

Divide both sides by \log(9).

$$-x+3=\frac{\log(81)}{\log(9)}$$

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

$$-x+3=\log_{9}\left(81\right)$$

Subtract 3 from both sides of the equation.

$$-x=2-3$$

Divide both sides by -1.

$$x=-\frac{1}{-1}$$

Solve for x (complex solution)

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

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