Solve e^1-5x=793

$${ e }^{ 1-5x } =793$$

Solve for x

$$x=\frac{1-\ln(793)}{5}\approx -1.135164644$$

Steps Using Definition of Logarithm

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

$$e^{-5x+1}=793$$

Take the logarithm of both sides of the equation.

$$\log(e^{-5x+1})=\log(793)$$

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

$$\left(-5x+1\right)\log(e)=\log(793)$$

Divide both sides by \log(e).

$$-5x+1=\frac{\log(793)}{\log(e)}$$

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

$$-5x+1=\log_{e}\left(793\right)$$

Subtract 1 from both sides of the equation.

$$-5x=\ln(793)-1$$

Divide both sides by -5.

$$x=\frac{\ln(793)-1}{-5}$$

Solve for x (complex solution)

$$x=-\frac{i\times 2\pi n_{1}}{5}-\frac{\ln(793)}{5}+\frac{1}{5}$$

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