Solve 32=2^3x-4

$$32 = 2 ^ { 3 x - 4 }$$

Solve for x

$$x=3$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

$$2^{3x-4}=32$$

Take the logarithm of both sides of the equation.

$$\log(2^{3x-4})=\log(32)$$

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

$$\left(3x-4\right)\log(2)=\log(32)$$

Divide both sides by \log(2).

$$3x-4=\frac{\log(32)}{\log(2)}$$

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

$$3x-4=\log_{2}\left(32\right)$$

Add 4 to both sides of the equation.

$$3x=5-\left(-4\right)$$

Divide both sides by 3.

$$x=\frac{9}{3}$$

Solve for x (complex solution)

$$x=\frac{i\times 2\pi n_{1}}{3\ln(2)}+3$$

$$n_{1}\in \mathrm{Z}$$