Solve 14^x+1=36

$$14 ^ { x + 1 } = 36$$

Solve for x

$$x=-\log_{14}\left(\frac{7}{18}\right)\approx 0.357878398$$

Steps Using Definition of Logarithm

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

$$14^{x+1}=36$$

Take the logarithm of both sides of the equation.

$$\log(14^{x+1})=\log(36)$$

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

$$\left(x+1\right)\log(14)=\log(36)$$

Divide both sides by \log(14).

$$x+1=\frac{\log(36)}{\log(14)}$$

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

$$x+1=\log_{14}\left(36\right)$$

Subtract 1 from both sides of the equation.

$$x=2\log_{14}\left(6\right)-1$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(14)}-\log_{14}\left(\frac{7}{18}\right)$$

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