Solve 2^3x*2^-4*2^6=?2

$$2 ^ { 3 x } \cdot 2 ^ { - 4 } \cdot 2 ^ { 6 } = ? 2$$

Solve for x

$$x=-\frac{1}{3}\approx -0.333333333$$

Steps Using Definition of Logarithm

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

$$4\times 2^{3x}=2$$

Divide both sides by 4.

$$2^{3x}=\frac{1}{2}$$

Take the logarithm of both sides of the equation.

$$\log(2^{3x})=\log(\frac{1}{2})$$

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

$$3x\log(2)=\log(\frac{1}{2})$$

Divide both sides by \log(2).

$$3x=\frac{\log(\frac{1}{2})}{\log(2)}$$

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

$$3x=\log_{2}\left(\frac{1}{2}\right)$$

Divide both sides by 3.

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

Solve for x (complex solution)

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

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