Solve (1/2)^2x+1=1

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

Steps Using Definition of Logarithm

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

$$\left(\frac{1}{2}\right)^{2x+1}=1$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{1}{2}\right)^{2x+1})=\log(1)$$

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

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

Divide both sides by \log(\frac{1}{2}).

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

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

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

Subtract 1 from both sides of the equation.

$$2x=-1$$

Divide both sides by 2.

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

Solve for x (complex solution)

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

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