Solve e^-3x+1=4

$${ e }^{ -3x+1 } =4$$

Solve for x

$$x=\frac{1-2\ln(2)}{3}\approx -0.128764787$$

Steps Using Definition of Logarithm

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

$$e^{-3x+1}=4$$

Take the logarithm of both sides of the equation.

$$\log(e^{-3x+1})=\log(4)$$

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

$$\left(-3x+1\right)\log(e)=\log(4)$$

Divide both sides by \log(e).

$$-3x+1=\frac{\log(4)}{\log(e)}$$

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

$$-3x+1=\log_{e}\left(4\right)$$

Subtract 1 from both sides of the equation.

$$-3x=2\ln(2)-1$$

Divide both sides by -3.

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

Solve for x (complex solution)

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

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