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
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Nice method
AOA Sir abhi tk hamara koi teacher set nai hoa
Which technique you used for 2^3 and compare
(x/y)^(x-y) = 2^3 is not necessary equivalent to x/y = 2 and x - y = 3. It is not rigorous.
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