Solve e^x+1=4

$$e ^ { x + 1 } = 4$$

Solve for x

$$x=2\ln(2)-1\approx 0.386294361$$

Steps Using Definition of Logarithm

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

$$e^{x+1}=4$$

Take the logarithm of both sides of the equation.

$$\log(e^{x+1})=\log(4)$$

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

$$\left(x+1\right)\log(e)=\log(4)$$

Divide both sides by \log(e).

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

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

$$x+1=\log_{e}\left(4\right)$$

Subtract 1 from both sides of the equation.

$$x=2\ln(2)-1$$

Solve for x (complex solution)

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

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