SOLUTION: log4 (4x+3) < log4 (5x-3)/(2x-3)

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Question 1186855: log4 (4x+3) < log4 (5x-3)/(2x-3)
Found 2 solutions by Edwin McCravy, mccravyedwin:
Answer by Edwin McCravy(20065) About Me  (Show Source):
Answer by mccravyedwin(409) About Me  (Show Source):
You can put this solution on YOUR website!
log4 (4x+3) < log4 (5x-3/2x-3)

I think you meant:

log%284%2C%284x%2B3%29%29+%3C+log%284%2C%28%285x-3%29%2F%282x-3%29%29%29

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

log%284%2C%28%285x-3%29%2F%282x-3%29%29%29+%3E+log%284%2C%284x%2B3%29%29

log%284%2C%285x-3%29%29-log%284%2C%282x-3%29%29+%3E+log%284%2C%284x%2B3%29%29

log%284%2C%285x-3%29%29-log%284%2C%282x-3%29%29+-+log%284%2C%284x%2B3%29%29%3E0

log%284%2C%28+++%285x-3%29%2F%28%282x-3%29%284x%2B3%29%29++++%29%29%3E0

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

%285x-3%29%2F%28%282x-3%29%284x%2B3%29%29%3E1

%285x-3%29%2F%28%282x-3%29%284x%2B3%29%29-1%3E0







%28-8x%5E2%2B11x%2B6%29%2F%28%282x-3%29%5E%22%22%284x%2B3%29%29%3E0

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



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%5E2%2B11x%2B6=0
8x%5E2-11x-6=0
x+=+%2811+%2B-+sqrt%28121-4%2A8%2A-6%29+%29%2F%282%2A8%29+
x+=+%2811+%2B-+sqrt%28313%29%29%2F16

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%2F4%3Cx%3C%2811+-+sqrt%28313%29%29%2F16
And, the graph is positive again when x is between the right-most
asymptote and the right-most x-intercept.

That's 3%2F2%3Cx%3C%2811+%2B+sqrt%28313%29%29%2F16

Answer:  -3%2F4%3Cx%3C%2811+-+sqrt%28313%29%29%2F16 and 3%2F2%3Cx%3C%2811+%2B+sqrt%28313%29%29%2F16

Edwin