## Intermediate Algebra (6th Edition)

$x=\frac{1}{3}(\frac{log(9)}{log(4)}-2)\approx-.1383$
According to the logarithm property of equality $log_{b}a=log_{b}c$ is equivalent to $a=c$ (where a, b, and c are real numbers such that $log_{b}a$ and $log_{b}c$ are real numbers and $b\ne1$). We can use this property to solve for x. $4^{3x+2}=9$ Take the common logarithm of both sides (which has base 10). $log(4^{3x+2})=log(9)$ Use the power property of logarithms. $(3x+2) log(4)=log(9)$ Divide both sides by $log(4)$. $3x+2=\frac{log(9)}{log(4)}$ Subtract 2 from both sides. $3x=\frac{log(9)}{log(4)}-2$ Divide both sides by 3. $x=\frac{1}{3}(\frac{log(9)}{log(4)}-2)\approx-.1383$