Solve 7^2x=1/9

$$7 ^ { 2 x } = \frac { 1 } { 9 }$$

Solve for x

$$x=-\log_{7}\left(3\right)\approx -0.564575034$$

Steps Using Definition of Logarithm

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

$$7^{2x}=\frac{1}{9}$$

Take the logarithm of both sides of the equation.

$$\log(7^{2x})=\log(\frac{1}{9})$$

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

$$2x\log(7)=\log(\frac{1}{9})$$

Divide both sides by \log(7).

$$2x=\frac{\log(\frac{1}{9})}{\log(7)}$$

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

$$2x=\log_{7}\left(\frac{1}{9}\right)$$

Divide both sides by 2.

$$x=-\frac{2\log_{7}\left(3\right)}{2}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(7)}-\log_{7}\left(3\right)$$

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