Solve 5.2^2x+1=320

$$5.2 ^ { 2 x + 1 } = 320$$

Solve for x

$$x=-\frac{\log_{5.2}\left(0.01625\right)}{2}\approx 1.24939824$$

Steps Using Definition of Logarithm

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

$$5.2^{2x+1}=320$$

Take the logarithm of both sides of the equation.

$$\log(5.2^{2x+1})=\log(320)$$

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

$$\left(2x+1\right)\log(5.2)=\log(320)$$

Divide both sides by \log(5.2).

$$2x+1=\frac{\log(320)}{\log(5.2)}$$

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

$$2x+1=\log_{5.2}\left(320\right)$$

Subtract 1 from both sides of the equation.

$$2x=\frac{\ln(320)}{\ln(\frac{26}{5})}-1$$

Divide both sides by 2.

$$x=\frac{\frac{\ln(320)}{\ln(\frac{26}{5})}-1}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{\ln(5.2)}-\frac{\log_{5.2}\left(0.01625\right)}{2}$$

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