Solve 5^2x-1=9

$${ 5 }^{ 2x-1 } =9$$

Solve for x

$$x=\log_{5}\left(3\right)+\frac{1}{2}\approx 1.182606194$$

Steps Using Definition of Logarithm

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

$$5^{2x-1}=9$$

Take the logarithm of both sides of the equation.

$$\log(5^{2x-1})=\log(9)$$

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

$$\left(2x-1\right)\log(5)=\log(9)$$

Divide both sides by \log(5).

$$2x-1=\frac{\log(9)}{\log(5)}$$

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

$$2x-1=\log_{5}\left(9\right)$$

Add 1 to both sides of the equation.

$$2x=2\log_{5}\left(3\right)-\left(-1\right)$$

Divide both sides by 2.

$$x=\frac{2\log_{5}\left(3\right)+1}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(5)}+\log_{5}\left(3\right)+\frac{1}{2}$$

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