How to solve 9 ^ (x + 1) = 81

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 9 ^ (x + 1) = 81 | Equatix

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$$9^{x+1}=81$$

$x=1$

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

$$9^{x+1}=81$$

Take the logarithm of both sides of the equation.

$$\log(9^{x+1})=\log(81)$$

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

$$\left(x+1\right)\log(9)=\log(81)$$

Divide both sides by $\log(9)$.

$$x+1=\frac{\log(81)}{\log(9)}$$

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

$$x+1=\log_{9}\left(81\right)$$

Subtract $1$ from both sides of the equation.

$$x=2-1$$

$x=\frac{i\pi n_{1}}{\ln(3)}+1$
$n_{1}\in \mathrm{Z}$