Solve frac{1}9^3x=81

$$\frac { 1 } { 9 } ^ { 3 x } = 81$$

Solve for x

$$x=-\frac{2}{3}\approx -0.666666667$$

Steps Using Definition of Logarithm

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

$$\left(\frac{1}{9}\right)^{3x}=81$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{9}\right)^{3x})=\log(81)$$

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

$$3x\log(\frac{1}{9})=\log(81)$$

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

$$3x=\frac{\log(81)}{\log(\frac{1}{9})}$$

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

$$3x=\log_{\frac{1}{9}}\left(81\right)$$

Divide both sides by 3.

$$x=-\frac{2}{3}$$

Solve for x (complex solution)

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

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