Solve 23^2x+1=54

$$23 ^ { 2 x + 1 } = 54$$

Solve for x

$$x=\frac{\log_{23}\left(54\right)-1}{2}\approx 0.136101324$$

Steps Using Definition of Logarithm

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

$$23^{2x+1}=54$$

Take the logarithm of both sides of the equation.

$$\log(23^{2x+1})=\log(54)$$

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

$$\left(2x+1\right)\log(23)=\log(54)$$

Divide both sides by \log(23).

$$2x+1=\frac{\log(54)}{\log(23)}$$

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

$$2x+1=\log_{23}\left(54\right)$$

Subtract 1 from both sides of the equation.

$$2x=\log_{23}\left(54\right)-1$$

Divide both sides by 2.

$$x=\frac{\log_{23}\left(54\right)-1}{2}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(23)}+\frac{\log_{23}\left(54\right)}{2}-\frac{1}{2}$$

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