A Nice Math Olympiad algebra exponential equation|

[@Entertainment-rn9vd]

Nice method

[@AbdulAhad-j1n2n]

AOA Sir abhi tk hamara koi teacher set nai hoa

[@FrazAhmad-n9m]

Which technique you used for 2^3 and compare

[@marcgriselhubert3915]

(x/y)^(x-y) = 2^3 is not necessary equivalent to x/y = 2 and x - y = 3. It is not rigorous.

[@noiha5721]

Exactly. You can do this, but then you have to prove that there are no more solutions. If x, y are natural, then there is at least one more solution x = 3, y = 6. If x, y are real, then there are infinitely many solutions, e.g. for any number a > 1, y = log(a)8 / (a-1), x = a*y. So e.g. a = 8, y = log(8)8 / 7 = 1/7, x = 8/7