Solve 25e^2x+1=962

$$25 e ^ { 2 x + 1 } = 962$$

Solve for x

$$x = \frac{\log_{e} {(\frac{962}{25})} - 1}{2} \approx 1.325069313$$

Steps Using Definition of Logarithm

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

$$25e^{2x+1}=962$$

Divide both sides by 25.

$$e^{2x+1}=\frac{962}{25}$$

Take the logarithm of both sides of the equation.

$$\log(e^{2x+1})=\log(\frac{962}{25})$$

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

$$\left(2x+1\right)\log(e)=\log(\frac{962}{25})$$

Divide both sides by \log(e).

$$2x+1=\frac{\log(\frac{962}{25})}{\log(e)}$$

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

$$2x+1=\log_{e}\left(\frac{962}{25}\right)$$

Subtract 1 from both sides of the equation.

$$2x=\ln(\frac{962}{25})-1$$

Divide both sides by 2.

$$x=\frac{\ln(\frac{962}{25})-1}{2}$$

Solve for x (complex solution)

$$x=\frac{\ln(\frac{962}{25})-1}{2}+\pi n_{1}i$$

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