Solve 3^2x-1=7^1+1

$$3 ^ { 2 x - 1 } = 7 ^ { 1 + 1 }$$

Solve for x

$$x=\log_{3}\left(7\right)+\frac{1}{2}\approx 2.271243749$$

Steps Using Definition of Logarithm

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

$$3^{2x-1}=49$$

Take the logarithm of both sides of the equation.

$$\log(3^{2x-1})=\log(49)$$

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

$$\left(2x-1\right)\log(3)=\log(49)$$

Divide both sides by \log(3).

$$2x-1=\frac{\log(49)}{\log(3)}$$

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

$$2x-1=\log_{3}\left(49\right)$$

Add 1 to both sides of the equation.

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

Divide both sides by 2.

$$x=\frac{2\log_{3}\left(7\right)+1}{2}$$

Solve for x (complex solution)

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

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