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℮^π or π^℮. Plz give a mathematical explanation for your answer.

2006-09-18 05:34:39 · 13 answers · asked by Swati M 1 in Science & Mathematics Mathematics

'e' is a mathematical constant which is used as the base of the natural logarithm.
The value of 'e' lies between 2 and 3.So,
e < π.

2006-09-18 05:54:17 · update #1

13 answers

let us take x^(1/x)
let y = x^(1/x)
ln y = ln x/x
x ln y = ln x
differentiate both sides
x/y dy/dx + ln y = 1/x
dy/dx = (1/x - ln y)/(x/y)
= (1/x-1/x ln x)/(x/y) = y/x^2(1-ln x)
for x > e this is -ve
so x^(1/x) decreases

so e^(1/e) > pi^(1/pi)
raise both to the power e * pi

e^pi > pi^ e

2006-09-18 06:34:04 · answer #1 · answered by Mein Hoon Na 7 · 2 1

Use scientific calculator or computer you will get e^pi > pi^e.
Numerical computation is the best mathematical explaination for any mathematical problem (if possible).

2006-09-19 23:27:39 · answer #2 · answered by Hemant 2 · 0 0

Hi:

I am by no means a math whiz but I have given it a shot.

Here goes...

The answer depends on the value of 'e'.

If 'e' is a positive whole number equal to or greater than 1, or when it is a decimal or a fraction , like 0.5, then pi^e is greater.

If it is a negative number, then the first expression is invalid.

Hope this answers your question.

Cheers!!

Hey, Jaques is right... just remembered that e is a constant

2006-09-18 12:47:51 · answer #3 · answered by BigPak 2 · 0 3

e ^pi is bigger.

Let's prove it.
e ^pi > pi ^e iff
pi log e > e log pi
iff
pi/log pi > e.
Look at the function f(x) = x/log(x).
By looking at its first derivative
(log x -1)/(log(x)^2), we see
that the function is decreasing for 0 < x < e,
has a minimum value at x = e and is increasing
for x >e. Since pi > e, pi/log pi > e,
so e^pi > pi ^e. (No calculator needed!)

2006-09-18 16:57:06 · answer #4 · answered by steiner1745 7 · 0 0

try to apply logarithm to the base e to both .then u will find that the log of e^pi is pi but for pi^e the value is 1.1447, which is less than pi hence e^pi is greater than pi^e.

2006-09-21 04:26:32 · answer #5 · answered by punith r 1 · 0 0

consider f(x) = ln (e^x/x^e)
= x - e*lnx
f'(x)= 1 -(e/x)
f'(x) = 0 implies x = e
f''(x) is - e/(x^2)
which is less than 0 for x = e
this implies max value of fn. is f(e) which is zero
so for any other x, f(x) will be less than zero
this means f(pi)<0

ln (e^pi/pi^e) < 0
and that means e^pi/pi^e <1

hence e^pi < pi^e

kp_math just chk ur differentiation....

2006-09-18 13:49:26 · answer #6 · answered by swajji 1 · 0 1

actually..therre is no meaning for the function e^pi....pi^e...they both are the syntax errors..
hence there values cannot be defined...
the value of pi=3.141592654 approx
and
e=2.71828

the gentleman who answererd above me is just a crap...
how can the value of an exponential function be negitive...
the min. value of all exponential functions is 0 at x=-infinity

2006-09-18 12:53:54 · answer #7 · answered by PIKACHU™ 3 · 0 2

e^pi = ~23.140693
pi^e = ~22.459158

to the other commenters, e is a mathematical constant
http://en.wikipedia.org/wiki/E_%28mathematical_constant%29

2006-09-18 12:42:26 · answer #8 · answered by Jaques S 3 · 2 0

Pie^e cause e is a variable that stands for any number so if you raised pie by a google then it would be a very big number but if you (google^pie) then it is a number almost 456.4654 as small

2006-09-18 12:41:18 · answer #9 · answered by nastynate 2 · 0 2

Assuming e is your variable then as long as e>pi pi^e would be larger.

2006-09-18 12:37:43 · answer #10 · answered by Sniper 4 · 0 2

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