How to solve 4 ^ 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 4 ^ x = 1/8 | Equatix

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$$4^{x}=\frac{1}{8}$$

$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$

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

$$4^{x}=\frac{1}{8}$$

Take the logarithm of both sides of the equation.

$$\log(4^{x})=\log(\frac{1}{8})$$

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

$$x\log(4)=\log(\frac{1}{8})$$

Divide both sides by $\log(4)$.

$$x=\frac{\log(\frac{1}{8})}{\log(4)}$$

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

$$x=\log_{4}\left(\frac{1}{8}\right)$$

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