Asked by gibbons1

solve the equation log3
(2-3x) = log9 (6x^2-19x+2)
Answered by Steve
since 9=3^2, log3 = 2*log9 and we have, using base 9 logs,

2log(2-3x) = log(6x^2-19x+2)
(2-3x)^2 = 6x^2 - 19x + 2
3x^2 + 7x + 2 = 0
(3x+1)(x+2) = 0
x = -2 or -1/3
Answered by sk
This answer is wrong
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