Solve 7 ^ (x + 2) - 2 = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$7^{x+2}-2=0$$

Solve for x

$x=\log_{7}\left(2\right)-2\approx -1.643792813$

Steps Using Definition of Logarithm

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

$$7^{x+2}-2=0$$

Add $2$ to both sides of the equation.

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

Take the logarithm of both sides of the equation.

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

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

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

Divide both sides by $\log(7)$.

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

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

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

Subtract $2$ from both sides of the equation.

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

Solve for x (complex solution)

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