Logarithmic inequality: 2*ln(x) - 3 < [ 2*ln(x) + 3 ] / ln(x)
- Thread starter Alberto02
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If I am reading the text in the attachments correctly, the exercise is as follows:
[imath]\qquad 2\ln(x) - 3 < \dfrac{2\ln(x)+3}{\ln(x)}[/imath]
You appear then to multiply through by [imath]\ln(x)[/imath]. However, the logarithm can take on negative values. You would need to consider two cases -- where [imath]0 < x < 1[/imath] and [imath]x > 1[/imath] -- before doing this multiplication.
However, one can achieve much the same result by converting to a common denominator. Then you'll have a quadratic inequality to solve.
(I get the same solution as is contained in "1.jpg". I don't follow what's going on in "2.jpg".)
Almost always, one of the first things I do after seeing one or more fractions is to consider clearing fractions.
[math]2 \ln(x) - 3 = \dfrac{2 \ln(x) + 3}{\ln(x)} \implies 2 \ln(x) * \ln(x) - 3 \ln (x) = 2 \ln(x) + 3[/math]
And when I see an ugly-looking expression multiplied by itself, I think u-substitution.
[math]u = \ln (x) \implies 2u^2 - 3u = 2u + 3 \implies \text{WHAT?}[/math]
This is a more mechanical way to see the same thing that stapel is seeing from experience.
[math]2 \ln(x) - 3 = \dfrac{2 \ln(x) + 3}{\ln(x)} \implies 2 \ln(x) * \ln(x) - 3 \ln (x) = 2 \ln(x) + 3[/math]
And when I see an ugly-looking expression multiplied by itself, I think u-substitution.
[math]u = \ln (x) \implies 2u^2 - 3u = 2u + 3 \implies \text{WHAT?}[/math]
This is a more mechanical way to see the same thing that stapel is seeing from experience.
Steven G
Elite Member
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...and what was your answer? Can you please post your work? If it's not 100% correct we can point out out your error(s) and if it is completely correct then someone else can benefit by seeing your work.
The helpers on this forum volunteer their free time helping students and all we ask is that students try 'paying' us back by making our life easier by not having to solve the same problems multiple time which is what may happen when a student does not show their completed work.
