Solve 9^2x=1/3^3

$$9 ^ { 2 x } = \frac { 1 } { 3 ^ { 3 } }$$

Solve for x

$$x=-\frac{3}{4}=-0.75$$

Steps Using Definition of Logarithm

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

$$9^{2x}=\frac{1}{27}$$

Take the logarithm of both sides of the equation.

$$\log(9^{2x})=\log(\frac{1}{27})$$

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

$$2x\log(9)=\log(\frac{1}{27})$$

Divide both sides by \log(9).

$$2x=\frac{\log(\frac{1}{27})}{\log(9)}$$

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

$$2x=\log_{9}\left(\frac{1}{27}\right)$$

Divide both sides by 2.

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

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{2\ln(3)}-\frac{3}{4}$$

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