SOLUTION: (a)(i)Solveforxintheequationlog(3x+2)−2logx=1−log(5x−3)[6mks]

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Question 1195766: (a)(i)Solveforxintheequationlog(3x+2)−2logx=1−log(5x−3)[6mks]
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39630) About Me  (Show Source):
You can put this solution on YOUR website!
The bunches of characters before and after are in the way.
log(3x+2)-2*log(x)=1-log(5x-3)
log%28%283x%2B2%29%29-2%2Alog%28%28x%29%29=1-log%28%285x%E2%88%923%29%29

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log%28%28%283x%2B2%29%285x-3%29%29%2Fx%5E2%29=1
put into exponential form assuming log is base 10,
10%5E1=%283x%2B2%29%285x-3%29%2Fx%5E2
10=%2815x%5E2%2B10x-9x-6%29%2Fx%5E2
10x%5E2=15x%5E2%2Bx-6
0=5x%5E2%2Bx-6, and you can finish.
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Answer by ikleyn(52914) About Me  (Show Source):
You can put this solution on YOUR website!
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Solve for x in the equation log(3x+2)−2logx = 1−log(5x−3)
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            (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

    %28%283x%2B2%29%2A%285x-3%29%29%2Fx%5E2 = 10.


It implies

    (3x+2)*(5x-3) = 10x^2.


Simplify

    5x^2 + x - 6 = 0.


Apply the quadratic formula.  The roots are

    x%5B1%2C2%5D = %28-1+%2B-+sqrt%28%28-1%29%5E2+-+4%2A5%2A%28-6%29%29%29%2F%282%2A5%29 = %28-1+%2B-+sqrt%28121%29%29%2F10 = %28-1+%2B-+11%29%2F10.


So the roots are  x%5B1%5D = 1  and/or  x%5B2%5D = -12%2F10 = -6%2F5.


Only root x= 1 is in the domain.


ANSWER.  The given equation has only one solution x = 1.

Solved.

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Ignore the post by @josgarithmetic,  since it is  WRONG  and  INCORRECT,
as the most part of his  " solutions "  at this forum.


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After seeing my post, @josgarithmetic fixed/rewrote his equation.