Question 1186855
<pre>
log4 (4x+3) < log4 (5x-3/2x-3)

I think you meant:

{{{log(4,(4x+3)) < log(4,((5x-3)/(2x-3)))}}}

Think I'll put the more complicated side on the left
and reverse the inequality:

{{{log(4,((5x-3)/(2x-3))) > log(4,(4x+3))}}}

{{{log(4,(5x-3))-log(4,(2x-3)) > log(4,(4x+3))}}}

{{{log(4,(5x-3))-log(4,(2x-3)) - log(4,(4x+3))>0}}}

{{{log(4,(   (5x-3)/((2x-3)(4x+3))    ))>0}}}

4 raised to the power of both sides will preserve the inequality
since log is an increasing function.

{{{(5x-3)/((2x-3)(4x+3))>1}}}

{{{(5x-3)/((2x-3)(4x+3))-1>0}}}

{{{(5x-3)/((2x-3)(4x+3))-((2x-3)(4x+3))/((2x-3)(4x+3))>0}}}

{{{(5x-3^"")/((2x-3)(4x+3)^"")-(8x^2-6x-9)/((2x-3)^""(4x+3))>0}}}

{{{(5x-3^"")/((2x-3)(4x+3)^"")+(-8x^2+6x+9)/((2x-3)^""(4x+3))>0}}}

{{{(-8x^2+11x+6)/((2x-3)^""(4x+3))>0}}}

It might help to draw a graph of y = the left side and see where the
graph is positive:

{{{drawing(400,400,-5,5,-5,5,
green(line(3/2,-6,3/2,6),line(-3/4,-6,-3/4,6)),
graph(400,400,-5,5,-5,5, (-8x^2+11x+6)/((2x-3)(4x+3))))}}}

It has vertical asymptotes at x=3/2 and x=-3/4
We need the x-intercepts which are the zeros of the numerator.

{{{-8x^2+11x+6=0}}}
{{{8x^2-11x-6=0}}}
{{{x = (11 +- sqrt(121-4*8*-6) )/(2*8) }}}
{{{x = (11 +- sqrt(313))/16}}}

So the graph is positive (above the x-axis) when x is between
the left horizontal asymptote and the left-most x-intercept,

That's {{{-3/4<x<(11 - sqrt(313))/16}}}
And, the graph is positive again when x is between the right-most
asymptote and the right-most x-intercept.

That's {{{3/2<x<(11 + sqrt(313))/16}}}

Answer:  {{{-3/4<x<(11 - sqrt(313))/16}}} and {{{3/2<x<(11 + sqrt(313))/16}}}

Edwin</pre>