How to solve 2 ^ (x - 1) = 32

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) = 32 | Equatix

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$$2^{x-1}=32$$

$x=6$

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

$$2^{x-1}=32$$

Take the logarithm of both sides of the equation.

$$\log(2^{x-1})=\log(32)$$

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

$$\left(x-1\right)\log(2)=\log(32)$$

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

$$x-1=\frac{\log(32)}{\log(2)}$$

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

$$x-1=\log_{2}\left(32\right)$$

Add $1$ to both sides of the equation.

$$x=5-\left(-1\right)$$

$x=\frac{i\times 2\pi n_{1}}{\ln(2)}+6$
$n_{1}\in \mathrm{Z}$