Solve e^-x=39

$$e ^ { - x } = 39$$

Solve for x

$$x=-\ln(39)\approx -3.663561646$$

Steps Using Definition of Logarithm

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

$$e^{-x}=39$$

Take the logarithm of both sides of the equation.

$$\log(e^{-x})=\log(39)$$

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

$$-x\log(e)=\log(39)$$

Divide both sides by \log(e).

$$-x=\frac{\log(39)}{\log(e)}$$

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

$$-x=\log_{e}\left(39\right)$$

Divide both sides by -1.

$$x=\frac{\ln(39)}{-1}$$

Solve for x (complex solution)

$$x=-2\pi n_{1}i-\ln(39)$$

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