IIT JEE , AIEEE Forum - Mathematics , Algebra

joyfrancis

^e > e....True/False?

chimanshu_007

^e > e^

= 3.14 , e = 2.7 (approx.) 1 way is put the values....

or

taking log

e log

>

loge

log e = 1

elog

>

now put the values....

u will get the answer

and i think the ans is false..

joyfrancis

we cannot just put the values , e is not exactly equal to 2.7 and pi is not 3.14. Can I have a general solution please.

amaron

Proofs for such problems can generally be given by Calculus

consider f(x)=x^(1/x)

f'(x) = (1/x)(x)^(1/x -1) - x^(1/x -2)(log x)

      =x^(1/x -2)[ 1- (log x)] <=0     for      x>=e

      therefore f decreases in[e,infinity)
      implies,

       (pi)^(1/pi) < e^(1/e), as pi>e
   implies,

        (pi)^(e) < (e)^(pi)

That solves the problem!

K4FECN6

We know that,

2<e<3.

Therefore, (pi)²<(pi)^e<(pi)^{3. ----->@}

But, e^pi=1+pi+[(pi)²/2]+[(pi)³/6]+.........

=6{ [1/6]+[(pi)/6]+[pi^squared/12]+[pi^cubed ] +.... ------>#

Therefore from @ and #, it is clear that e^pi>(pi)^e.

Please,rate me if I am correct.

aditya928

its a good question if only one can give a perfect represented answer....

spideyunlimited

take ln of pi^e and e^pi

e ln(pi) and (pi) lne

as ln e = 1
e ln(pi) and pi

ln pi will be approx near 1.2 in value as e to power someting should be pi so approx 1.2 can be taken.
e (1.2) and pi
2.7(1.2) and 3.14 approx.s

therefore right hand qty .is greater... therefore e^pi > pi^e and wht u have mentioned ( given statement) is false

Ask & Discuss Questions with Community & Experts