solve the inequality e^x - ln(x) <= e/x for x

Hussein Ali 99 said:
solve the inequality #e^x-lnx<=e/x#
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Hussein Ali 99 said:
solve the inequality #e^x-lnx<=e/x#
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Before we can try to help, we need to be sure of the question. Do you mean, as what you wrote means, e^x - ln(x) <= e/x, that is, \(\displaystyle e^x - \ln(x) \le \frac{e}{x}\), or e^(x - ln(x)) <= e/x, that is, \(\displaystyle e^{x - \ln(x)} \le \frac{e}{x}\)? The latter, I think, will be much easier.
 
it is \(\displaystyle e^{x - \ln(x)} \le \frac{e}{x}\)

i am sure that the solution [0<x<=1] but i need the proof.
 
Hussein Ali 99 said:
it is \(\displaystyle e^{x - \ln(x)} \le \frac{e}{x}\)

i am sure that the solution [0<x<=1] but i need the proof.
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First step - simplify e^ln(x) = ??

Second step - What is the domain of the expression above?

continue....
 
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Solve the inequality:

. . . . .\(\displaystyle e^x\, -\, \ln(x)\, \leq\, \frac{e}{x}\)

i am sure that the solution is the interval {0<x<=1}, but i need the proof.
 
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Hussein Ali 99 said:
Solve the inequality:

. . . . .\(\displaystyle e^x\, -\, \ln(x)\, \leq\, \frac{e}{x}\)

i am sure that the solution is the interval {0<x<=1}, but i need the proof.
Click to expand...
Well, yes, we can certainly be fairly certain of the solution, since this can be found online.

You posted this to "Calculus". Please provide a listing of recent topics of study in your calculus class (in particular, the topic for the section for which this is a homework exercise), as this will be helpful in determining how you're expected to solve this. When you reply, please include a clear listing of your thoughts and efforts so far. Thank you! ;)
 
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