# How do you solve log _ 6 (2-x) + log _ 6 (3-4x) = 1?

Using the property ${\log}_{b} a + {\log}_{b} c = {\log}_{b} a c$ , we can rewrite the equation as -
${\log}_{6} \left(2 - x\right) \left(3 - 4 x\right) = 1$
Solving this we get x=0 or x=$\frac{11}{4}$