Solve 3^x+2=7^5-3

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

Solve for x

$$x=2\left(\log_{3}\left(7\right)-1\right)\approx 1.542487498$$

Steps Using Definition of Logarithm

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

$$3^{x+2}=49$$

Take the logarithm of both sides of the equation.

$$\log(3^{x+2})=\log(49)$$

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

$$\left(x+2\right)\log(3)=\log(49)$$

Divide both sides by \log(3).

$$x+2=\frac{\log(49)}{\log(3)}$$

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

$$x+2=\log_{3}\left(49\right)$$

Subtract 2 from both sides of the equation.

$$x=2\log_{3}\left(7\right)-2$$

Solve for x (complex solution)

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

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