Solve 2.3^2x+1=54

$$2.3 ^ { 2 x + 1 } = 54$$

Solve for x

$$x=-\frac{\log_{2.3}\left(\frac{23}{540}\right)}{2}\approx 1.89460941$$

Steps Using Definition of Logarithm

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

$$2.3^{2x+1}=54$$

Take the logarithm of both sides of the equation.

$$\log(2.3^{2x+1})=\log(54)$$

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

$$\left(2x+1\right)\log(2.3)=\log(54)$$

Divide both sides by \log(2.3).

$$2x+1=\frac{\log(54)}{\log(2.3)}$$

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

$$2x+1=\log_{2.3}\left(54\right)$$

Subtract 1 from both sides of the equation.

$$2x=\frac{\ln(54)}{\ln(\frac{23}{10})}-1$$

Divide both sides by 2.

$$x=\frac{\frac{\ln(54)}{\ln(\frac{23}{10})}-1}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(2.3)}-\frac{\log_{2.3}\left(\frac{23}{540}\right)}{2}$$

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