Question

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

Answer 1

Using the property `log_b a + log_b c = log_b ac` , we can rewrite the equation as -

`log_6 (2-x)(3-4x) = 1`

This is the same as saying (2-x)(3-4x)=6

Solving this we get x=0 or x=`11/4`