Solve 3^x+1=2000

$${ 3 }^{ x+1 } =2000$$

Solve for x

$$x=\log_{3}\left(2000\right)-1\approx 5.918639576$$

Steps Using Definition of Logarithm

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

$$3^{x+1}=2000$$

Take the logarithm of both sides of the equation.

$$\log(3^{x+1})=\log(2000)$$

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

$$\left(x+1\right)\log(3)=\log(2000)$$

Divide both sides by \log(3).

$$x+1=\frac{\log(2000)}{\log(3)}$$

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

$$x+1=\log_{3}\left(2000\right)$$

Subtract 1 from both sides of the equation.

$$x=\log_{3}\left(2000\right)-1$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(3)}+\log_{3}\left(2000\right)-1$$

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