Solve 12^2x+3=149

$${ 12 }^{ 2x+3 } = 149$$

Solve for x

$$x=\frac{\log_{12}\left(149\right)-3}{2}\approx -0.493131934$$

Steps Using Definition of Logarithm

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

$$12^{2x+3}=149$$

Take the logarithm of both sides of the equation.

$$\log(12^{2x+3})=\log(149)$$

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

$$\left(2x+3\right)\log(12)=\log(149)$$

Divide both sides by \log(12).

$$2x+3=\frac{\log(149)}{\log(12)}$$

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

$$2x+3=\log_{12}\left(149\right)$$

Subtract 3 from both sides of the equation.

$$2x=\log_{12}\left(149\right)-3$$

Divide both sides by 2.

$$x=\frac{\log_{12}\left(149\right)-3}{2}$$

Solve for x (complex solution)

$$x=\frac{\pi n_{1}i}{\ln(12)}+\frac{\log_{12}\left(149\right)}{2}-\frac{3}{2}$$

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