Solve 2^2x-2=1/16

$$2 ^ { 2 x - 2 } = \frac { 1 } { 16 }$$

Solve for x

$$x=-1$$

Steps Using Definition of Logarithm

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

$$2^{2x-2}=\frac{1}{16}$$

Take the logarithm of both sides of the equation.

$$\log(2^{2x-2})=\log(\frac{1}{16})$$

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

$$\left(2x-2\right)\log(2)=\log(\frac{1}{16})$$

Divide both sides by \log(2).

$$2x-2=\frac{\log(\frac{1}{16})}{\log(2)}$$

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

$$2x-2=\log_{2}\left(\frac{1}{16}\right)$$

Add 2 to both sides of the equation.

$$2x=-4-\left(-2\right)$$

Divide both sides by 2.

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