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If the function f:(0,infinite)-->R, f(x)=lnx/x proove that e^pi>pi^e

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lightjazz | Student, Undergraduate | eNoter

Posted June 25, 2012 at 3:19 PM via web

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If the function f:(0,infinite)-->R, f(x)=lnx/x proove that e^pi>pi^e

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted June 25, 2012 at 4:08 PM (Answer #1)

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You would find very helpful if you would know about the extrema of the function and its monotony.

You should solve the equation f'(x)=0 to find the maximum or minimum value of the function such that:

`f'(x) = (1/x*x - lnx)/(x^2) `

`f'(x) = (1-ln x)/(x^2)`

You need to solve for x the equation f'(x) = 0 such that:

`1-ln x = 0 =gt ln x = 1 =gt x = e`

You need to evaluate f(e) such that:

`f(e) = (ln e)/e =gt f(e) = 1/e`

Notice that the derivative is positive for `x in (1,e)`  and it is negative for `x in (e,oo), ` hence the function reaches its maximum at f(e).

Since the function decreases over `(e,oo), ` hence for `eltpi =gt f(e)gtf(pi).`

You shoud substitute `e`  and `pi`  for x in equation of the function such that:

`1/e gt (ln pi)/pi =gt pi gt e*ln pi`

You may write the inequality in the following way:

`pi*1 gt e*ln pi`

Substituting ln e for 1 yields:

`pi*ln e gt e*ln pi`

Using the properties of logarithms yields:

`ln e^pi gt ln pi^e`

Since the base of logarithm is larger than 1, then `e^pi gt pi^e` .

Hence, using monotony of the given function (the function decreases over `(e,oo))` yields that `e^pi gt pi^e` .

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