Solve 2^3x*4^-1=16

$$2 ^ { 3 x } \times 4 ^ { - 1 } = 16$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

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

$$\frac{1}{4}\times 2^{3x}=16$$

Multiply both sides by 4.

$$2^{3x}=64$$

Take the logarithm of both sides of the equation.

$$\log(2^{3x})=\log(64)$$

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

$$3x\log(2)=\log(64)$$

Divide both sides by \log(2).

$$3x=\frac{\log(64)}{\log(2)}$$

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

$$3x=\log_{2}\left(64\right)$$

Divide both sides by 3.

$$x=\frac{6}{3}$$

Solve for x (complex solution)

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

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