How to solve (2)^(x+1)=8

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 (2)^(x+1)=8 | Equatix

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$${ 2 }^{ x+1 } =8$$

$x=2$

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

$$2^{x+1}=8$$

Take the logarithm of both sides of the equation.

$$\log(2^{x+1})=\log(8)$$

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

$$\left(x+1\right)\log(2)=\log(8)$$

Divide both sides by $\log(2)$.

$$x+1=\frac{\log(8)}{\log(2)}$$

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

$$x+1=\log_{2}\left(8\right)$$

Subtract $1$ from both sides of the equation.

$$x=3-1$$

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