Solve (1/9)^x-2=1/9

$$( \frac { 1 } { 9 } ) ^ { x - 2 } = \frac { 1 } { 9 }$$

Solve for x

$$x=3$$

Steps Using Definition of Logarithm

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

$$\left(\frac{1}{9}\right)^{x-2}=\frac{1}{9}$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{9}\right)^{x-2})=\log(\frac{1}{9})$$

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

$$\left(x-2\right)\log(\frac{1}{9})=\log(\frac{1}{9})$$

Divide both sides by \log(\frac{1}{9}).

$$x-2=\frac{\log(\frac{1}{9})}{\log(\frac{1}{9})}$$

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

$$x-2=\log_{\frac{1}{9}}\left(\frac{1}{9}\right)$$

Add 2 to both sides of the equation.

$$x=1-\left(-2\right)$$

Solve for x (complex solution)

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

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