Solve 2^2x=1/8

$${ 2 }^{ 2x } = \frac{ 1 }{ 8 }$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

Steps Using Definition of Logarithm

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

$$2^{2x}=\frac{1}{8}$$

Take the logarithm of both sides of the equation.

$$\log(2^{2x})=\log(\frac{1}{8})$$

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

$$2x\log(2)=\log(\frac{1}{8})$$

Divide both sides by \log(2).

$$2x=\frac{\log(\frac{1}{8})}{\log(2)}$$

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

$$2x=\log_{2}\left(\frac{1}{8}\right)$$

Divide both sides by 2.

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

Solve for x (complex solution)

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

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