Solve 2^x+1+2^x+1=20

$${ 2 }^{ x+1 } + { 2 }^{ x+1 } = 20$$

Solve for x

$$x=\log_{2}\left(5\right)\approx 2.321928095$$

Steps Using Definition of Logarithm

Combine 2^{x+1} and 2^{x+1} to get 2\times 2^{x+1}.

$$2\times 2^{x+1}=20$$

Divide both sides by 2.

$$2^{x+1}=\frac{20}{2}$$

Divide 20 by 2 to get 10.

$$2^{x+1}=10$$

Take the logarithm of both sides of the equation.

$$\log(2^{x+1})=\log(10)$$

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

$$\left(x+1\right)\log(2)=\log(10)$$

Divide both sides by \log(2).

$$x+1=\frac{\log(10)}{\log(2)}$$

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

$$x+1=\log_{2}\left(10\right)$$

Subtract 1 from both sides of the equation.

$$x=\log_{2}\left(10\right)-1$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(2)}+\log_{2}\left(5\right)$$

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