Question

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!

Answer 1

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.