Solve 8^x*2=128

$$8 ^ { x } \cdot 2 = 128$$

Solve for x

$$x=2$$

Steps Using Definition of Logarithm

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

$$2\times 8^{x}=128$$

Divide both sides by 2.

$$8^{x}=64$$

Take the logarithm of both sides of the equation.

$$\log(8^{x})=\log(64)$$

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

$$x\log(8)=\log(64)$$

Divide both sides by \log(8).

$$x=\frac{\log(64)}{\log(8)}$$

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

$$x=\log_{8}\left(64\right)$$

Solve for x (complex solution)

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

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