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) log3(2−3x)=log9(6x2−19x−2)

Thanks!

1 Answer
Ananda Dasgupta
Feb 26, 2018

There are no real solutions to this problem

Explanation:

Using log_a b = log a/log blogab=logalogb we can rewrite the equation
log_3(2-3x) = log_9(6x^2-19x-2)log3(2−3x)=log9(6x2−19x−2)
as

log(2-3x)/log3=log(6x^2-19x-2)/log9 implies log(2−3x)log3=log(6x2−19x−2)log9⇒
log(2-3x)times log9/log3=log(6x^2-19x-2) implies log(2−3x)×log9log3=log(6x2−19x−2)⇒
2 log(2-3x) =log(6x^2-19x-2) implies 2log(2−3x)=log(6x2−19x−2)⇒
log(2-3x)^2 =log(6x^2-19x-2) implies log(2−3x)2=log(6x2−19x−2)⇒
(2-3x)^2 =6x^2-19x-2 implies(2−3x)2=6x2−19x−2⇒
4-12x+9x^2 = 6x^2-19x-2 implies4−12x+9x2=6x2−19x−2⇒
3x^2+7x+6=03x2+7x+6=0

This quadratic equation has a discriminant of
7^2-4times 3 times 6 = 49-72 = -23<072−4×3×6=49−72=−23<0
so there are no real roots.

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