Solve 2^x+1+2^2*-1=10

$${ 2 }^{ x+1 } + { 2 }^{ 2 \times -1 } =10$$

Solve for x

$$x=\log_{2}\left(39\right)-3\approx 2.285402219$$

Steps Using Definition of Logarithm

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

$$2^{x+1}+\frac{1}{4}=10$$

Subtract \frac{1}{4} from both sides of the equation.

$$2^{x+1}=\frac{39}{4}$$

Take the logarithm of both sides of the equation.

$$\log(2^{x+1})=\log(\frac{39}{4})$$

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

$$\left(x+1\right)\log(2)=\log(\frac{39}{4})$$

Divide both sides by \log(2).

$$x+1=\frac{\log(\frac{39}{4})}{\log(2)}$$

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

$$x+1=\log_{2}\left(\frac{39}{4}\right)$$

Subtract 1 from both sides of the equation.

$$x=\log_{2}\left(39\right)-2-1$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(2)}+\log_{2}\left(39\right)-3$$

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