since the logs are equal,
(2-3x)^3 = (6x^2-19x+2)^9
x = 0.04086, 2.94615
Somehow I doubt that's what you had in mind.
If you meant that the first log is base 3, and the second is base 9, then maybe you should have explained things a bit. Many people would write
log(base 3)(2-3x) as log_3(2-3x) so the underscore indicates the base.
In any case, if that's what you meant, then since 9=3^2, log_3(x) = 2log_9(x). That would give you
2log_9(2-3x) = log_9(6x^2-19x+2)
(2-3x)^2 = 6x^2-19x+2
3x^2 + 7x + 2 = 0
(3x+1)(x+2) = 0
x = -1/3, -2
Now all you need to do is check to make sure that those roots satisfy the original equation. They need to be positive for both logs.
log (2-3x)^3=log (6x^2-19x+2)^9
2 answers
"anti-log" both sides
(2-3x)^3 = 6x^2 - 19x + 2)^9
take cube root of both sides
2-3x = (6x^2 - 19x + 2)^3
Was hoping that 6x^2 - 19x + 2 would factor but it doesn't, so what a mess !!.....
you will need "technology" to solve this equation.
I let Wolfram do it:
Got 2 real roots, but one of them takes the log of a negative.
So only x = .0408623.. only will work
https://www.wolframalpha.com/input/?i=solve+2-3x+%3D+%286x%5E2+-+19x+%2B+2%29%5E3
(2-3x)^3 = 6x^2 - 19x + 2)^9
take cube root of both sides
2-3x = (6x^2 - 19x + 2)^3
Was hoping that 6x^2 - 19x + 2 would factor but it doesn't, so what a mess !!.....
you will need "technology" to solve this equation.
I let Wolfram do it:
Got 2 real roots, but one of them takes the log of a negative.
So only x = .0408623.. only will work
https://www.wolframalpha.com/input/?i=solve+2-3x+%3D+%286x%5E2+-+19x+%2B+2%29%5E3