Solve 40e^x-1=20

$$40 { e }^{ x-1 } =20$$

Solve for x

$$x=1-\ln(2)\approx 0.306852819$$

Steps Using Definition of Logarithm

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

$$40e^{x-1}=20$$

Divide both sides by 40.

$$e^{x-1}=\frac{1}{2}$$

Take the logarithm of both sides of the equation.

$$\log(e^{x-1})=\log(\frac{1}{2})$$

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

$$\left(x-1\right)\log(e)=\log(\frac{1}{2})$$

Divide both sides by \log(e).

$$x-1=\frac{\log(\frac{1}{2})}{\log(e)}$$

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

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

Add 1 to both sides of the equation.

$$x=-\ln(2)-\left(-1\right)$$

Solve for x (complex solution)

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

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