Solve 8e^-2x-16=0

$$8 e ^ { - 2 x } - 16 = 0$$

Solve for x

$$x=-\frac{\ln(2)}{2}\approx -0.34657359$$

Steps Using Definition of Logarithm

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

$$8e^{-2x}-16=0$$

Add 16 to both sides of the equation.

$$8e^{-2x}=16$$

Divide both sides by 8.

$$e^{-2x}=2$$

Take the logarithm of both sides of the equation.

$$\log(e^{-2x})=\log(2)$$

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

$$-2x\log(e)=\log(2)$$

Divide both sides by \log(e).

$$-2x=\frac{\log(2)}{\log(e)}$$

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

$$-2x=\log_{e}\left(2\right)$$

Divide both sides by -2.

$$x=\frac{\ln(2)}{-2}$$

Solve for x (complex solution)

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

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