## Algebra 2 (1st Edition)

$$x=\frac{5\ln \left(6\right)+4\ln \left(3\right)}{\ln \left(12\right)}$$
Using the rules of logarithms and exponents, we find: $$\ln \left(3^{x+4}\right)=\ln \left(6^{2x-5}\right)\\ \left(x+4\right)\ln \left(3\right)=\left(2x-5\right)\ln \left(6\right) \\ x=\frac{5\ln \left(6\right)+4\ln \left(3\right)}{\ln \left(12\right)}$$