Solve 5e^2x+3+3=10

$$5 { e }^{ 2x+3 } +3 = 10$$

Solve for x

$$x=\frac{\ln(\frac{7}{5})-3}{2}\approx -1.331763882$$

Steps Using Definition of Logarithm

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

$$5e^{2x+3}+3=10$$

Subtract 3 from both sides of the equation.

$$5e^{2x+3}=7$$

Divide both sides by 5.

$$e^{2x+3}=\frac{7}{5}$$

Take the logarithm of both sides of the equation.

$$\log(e^{2x+3})=\log(\frac{7}{5})$$

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

$$\left(2x+3\right)\log(e)=\log(\frac{7}{5})$$

Divide both sides by \log(e).

$$2x+3=\frac{\log(\frac{7}{5})}{\log(e)}$$

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

$$2x+3=\log_{e}\left(\frac{7}{5}\right)$$

Subtract 3 from both sides of the equation.

$$2x=\ln(\frac{7}{5})-3$$

Divide both sides by 2.

$$x=\frac{\ln(\frac{7}{5})-3}{2}$$

Solve for x (complex solution)

$$x=\frac{\ln(\frac{7}{5})-3}{2}+\pi n_{1}i$$

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