Solve 2^-x=40

$$2 ^ { - x } = 40$$

Solve for x

$$x=-\log_{2}\left(40\right)\approx -5.321928095$$

Steps Using Definition of Logarithm

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

$$2^{-x}=40$$

Take the logarithm of both sides of the equation.

$$\log(2^{-x})=\log(40)$$

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

$$-x\log(2)=\log(40)$$

Divide both sides by \log(2).

$$-x=\frac{\log(40)}{\log(2)}$$

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

$$-x=\log_{2}\left(40\right)$$

Divide both sides by -1.

$$x=\frac{\log_{2}\left(40\right)}{-1}$$

Solve for x (complex solution)

$$x=-\frac{2\pi n_{1}i}{\ln(2)}-\log_{2}\left(5\right)-3$$

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