Solve (1/9)^x=9

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

Solve for x

$$x=-1$$

Steps Using Definition of Logarithm

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

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

Take the logarithm of both sides of the equation.

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

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

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

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

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

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

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

Solve for x (complex solution)

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

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