.
Solve for x in the equation log(3x+2)−2logx = 1−log(5x−3)
~~~~~~~~~~~~~~~~~~~~
(1) See how I edited your post to make it readable.
Are you familiar with the conception of the blank symbol,
which is used between the words to make reading possible ?
(2) In his post, @josgarithmetic reduced the problem
to incorrect equation 20x^2 + x - 6 = 0, which is IRRELEVANT to the problem.
So I came to bring a correct solution.
Your starting equation is
log(3x+2)−2logx = 1−log(5x−3)
The domain of this equation (where logarithms are defined) is
x > -3/2; x > 0; x > 3/5, which lead to the final inequality x > 3/5.
From the equation
= 10.
It implies
(3x+2)*(5x-3) = 10x^2.
Simplify
5x^2 + x - 6 = 0.
Apply the quadratic formula. The roots are
= = = .
So the roots are = 1 and/or = = .
Only root x= 1 is in the domain.
ANSWER. The given equation has only one solution x = 1.
Solved.
-----------------
Ignore the post by @josgarithmetic, since it is WRONG and INCORRECT,
as the most part of his " solutions " at this forum.
///////////////
After seeing my post, @josgarithmetic fixed/rewrote his equation.