Solve 2^2x+3=8

$$2 ^ { 2 x + 3 } = 8$$

Solve for x

$$x=0$$

Steps Using Definition of Logarithm

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

$$2^{2x+3}=8$$

Take the logarithm of both sides of the equation.

$$\log(2^{2x+3})=\log(8)$$

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

$$\left(2x+3\right)\log(2)=\log(8)$$

Divide both sides by \log(2).

$$2x+3=\frac{\log(8)}{\log(2)}$$

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

$$2x+3=\log_{2}\left(8\right)$$

Subtract 3 from both sides of the equation.

$$2x=3-3$$

Divide both sides by 2.

$$x=\frac{0}{2}$$

Solve for x (complex solution)

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

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