Solve 5e^2x-7=11

$$5 e ^ { 2 x - 7 } = 11$$

Solve for x

$$x = \frac{\log_{e} {(\frac{11}{5})} + 7}{2} \approx 3.89422868$$

Steps Using Definition of Logarithm

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

$$5e^{2x-7}=11$$

Divide both sides by 5.

$$e^{2x-7}=\frac{11}{5}$$

Take the logarithm of both sides of the equation.

$$\log(e^{2x-7})=\log(\frac{11}{5})$$

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

$$\left(2x-7\right)\log(e)=\log(\frac{11}{5})$$

Divide both sides by \log(e).

$$2x-7=\frac{\log(\frac{11}{5})}{\log(e)}$$

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

$$2x-7=\log_{e}\left(\frac{11}{5}\right)$$

Add 7 to both sides of the equation.

$$2x=\ln(\frac{11}{5})-\left(-7\right)$$

Divide both sides by 2.

$$x=\frac{\ln(\frac{11}{5})+7}{2}$$

Solve for x (complex solution)

$$x=\frac{\ln(\frac{11}{5})+7}{2}+\pi n_{1}i$$

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