Solve 243=9^2x+1

$$243 = 9 ^ { 2 x + 1 }$$

Solve for x

$$x=\frac{3}{4}=0.75$$

Steps Using Definition of Logarithm

Swap sides so that all variable terms are on the left hand side.

$$9^{2x+1}=243$$

Take the logarithm of both sides of the equation.

$$\log(9^{2x+1})=\log(243)$$

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

$$\left(2x+1\right)\log(9)=\log(243)$$

Divide both sides by \log(9).

$$2x+1=\frac{\log(243)}{\log(9)}$$

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

$$2x+1=\log_{9}\left(243\right)$$

Subtract 1 from both sides of the equation.

$$2x=\frac{5}{2}-1$$

Divide both sides by 2.

$$x=\frac{\frac{3}{2}}{2}$$

Solve for x (complex solution)

$$x=\frac{i\pi n_{1}}{2\ln(3)}+\frac{3}{4}$$

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