Question

Solve the following equations
x^a^x=a^x^a, knowing that a<1 and a>0.
I love this exercise!

Answer 1

x^(ax)= a^(ax) ==> ln [x^(ax)]= ln [a^(ax)] ===> ax *ln x= ax ln a ===> ln x = ln a ===> x=a

Answer 2

Nope. BEWARE! It is not x^(ax). It is x^a^x. NOT THE SAME THING!!I wouldn't post a easy question!
\[x^{a^{x}}=a^{x^{a}}\]

Answer 3

it's really a matter of notations writing method online, you maight be right. Thank you.

Answer 4

in your case: x^(a^x) = a^(x^a) , this is the conventional way to write the experession withouth math editor

Answer 5

not as been posted at the begning

Answer 6

My mistake :)

Answer 7

x=a is still a solution.

it should be obvious that:
\[a^{a^a}=a^{a^a}\]

Answer 8

Yes, it is true! x=a is a solution, indeed, but one has to show that it is unique. And this is the nice part of the problem....

Answer 9

ok, great then to make such a momentum.

Answer 10

Good luck with it!