Solve 7^x+2=19

$$7 ^ { x + 2 } = 19$$

Solve for x

$$x=\log_{7}\left(19\right)-2\approx -0.486857689$$

Steps Using Definition of Logarithm

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

$$7^{x+2}=19$$

Take the logarithm of both sides of the equation.

$$\log(7^{x+2})=\log(19)$$

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

$$\left(x+2\right)\log(7)=\log(19)$$

Divide both sides by \log(7).

$$x+2=\frac{\log(19)}{\log(7)}$$

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

$$x+2=\log_{7}\left(19\right)$$

Subtract 2 from both sides of the equation.

$$x=\log_{7}\left(19\right)-2$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{\ln(7)}+\log_{7}\left(19\right)-2$$

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