e-pi-gt-pi-e-or-e-pi-lt-pi-e-

Question Number 32897 by artibunja last updated on 06/Apr/18

e^{π} > π^{e} o r e^{π} < π^{e} ?

Answered by MJS last updated on 06/Apr/18

e^{x} = x^{e}; x > 1 \Rightarrow x = e \Rightarrow \forall x > e : e^{x} > x^{e}

π > e \Rightarrow e^{π} > π^{e}

Commented by MJS last updated on 06/Apr/18

interestingly for a^{x} = x^{a} with a > 1

and x > 1 two solutions seem to

exist (i . e . 2^{x} = x^{2} \Rightarrow x = 2 \lor x = 4;

3^{x} = x^{3} \Rightarrow x = 3 \lor x \approx 2.47805) but

for a = e {there}^{'} s only one : x = e

can anybody further explain this ?

Commented by mrW2 last updated on 06/Apr/18

s o l u t i o n f o r a^{x} = x^{a} (a > 0) :

a^{\frac{x}{a}} = x

e^{\frac{x}{a} \ln a} = x

(- \frac{x}{a} \ln a) e^{- \frac{x}{a} \ln a} = - \frac{\ln a}{a}

\Rightarrow - \frac{x}{a} \ln a = W (- \frac{\ln a}{a})

\Rightarrow x = - \frac{a}{\ln a} W (- \frac{\ln a}{a}) = - \frac{W (- \ln a^{\frac{1}{a}})}{\ln a^{\frac{1}{a}}}

i f a = e : o n e s o l u t i o n

i f a \neq e : t w o s o l u t i o n s

Answered by Tinkutara last updated on 06/Apr/18

C o n s i d e r f (x) = y = x^{\frac{1}{x}}

\frac{d y}{d x} = \frac{y}{x^{2}} (1 - \ln x)

I f x < e, f (x) i s i n c r e a s i n g .

I f x > e, f (x) i s d e c r e a s i n g .

∴ e^{\frac{1}{e}} > π^{\frac{1}{π}}

e^{π} > π^{e}

Answered by artibunja last updated on 06/Apr/18