Working with logarithms?

I know the log rules and I am comfortable changing bases and working with log equations, but this question is confusing:

log_3(2-3x)=log_9(6x^2-19x-2)

Thanks!

1 Answer
Feb 26, 2018

There are no real solutions to this problem

Explanation:

Using log_a b = log a/log b we can rewrite the equation
log_3(2-3x) = log_9(6x^2-19x-2)
as

log(2-3x)/log3=log(6x^2-19x-2)/log9 implies
log(2-3x)times log9/log3=log(6x^2-19x-2) implies
2 log(2-3x) =log(6x^2-19x-2) implies
log(2-3x)^2 =log(6x^2-19x-2) implies
(2-3x)^2 =6x^2-19x-2 implies
4-12x+9x^2 = 6x^2-19x-2 implies
3x^2+7x+6=0

This quadratic equation has a discriminant of
7^2-4times 3 times 6 = 49-72 = -23<0
so there are no real roots.