How to solve 7 ^ (x + 3) = 1

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve 7 ^ (x + 3) = 1 | Equatix

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$$7^{x+3}=1$$

$x=-3$

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

$$7^{x+3}=1$$

Take the logarithm of both sides of the equation.

$$\log(7^{x+3})=\log(1)$$

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

$$\left(x+3\right)\log(7)=\log(1)$$

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

$$x+3=\frac{\log(1)}{\log(7)}$$

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

$$x+3=\log_{7}\left(1\right)$$

Subtract $3$ from both sides of the equation.

$$x=-3$$

$x=\frac{i\times 2\pi n_{1}}{\ln(7)}-3$
$n_{1}\in \mathrm{Z}$