Solve 7^x+1=88

$$7 ^ { x + 1 } = 88$$

Solve for x

$$x=\log_{7}\left(88\right)-1\approx 1.300895967$$

Steps Using Definition of Logarithm

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

$$7^{x+1}=88$$

Take the logarithm of both sides of the equation.

$$\log(7^{x+1})=\log(88)$$

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

$$\left(x+1\right)\log(7)=\log(88)$$

Divide both sides by \log(7).

$$x+1=\frac{\log(88)}{\log(7)}$$

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

$$x+1=\log_{7}\left(88\right)$$

Subtract 1 from both sides of the equation.

$$x=\log_{7}\left(88\right)-1$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(7)}+\log_{7}\left(88\right)-1$$

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