If x^a * x^b = (x^a)^b, prove that a/b + b/a = -1.
Best I can get is a/b + b/a = ab - 2. Is there any way to show ab = 1 or other way to approach the initial problem to find the proof?
from the first equation:
x^a * x^b = (x^a)^b
x^(a+b) = x^(ab)
thus a+b = ab
the only integer solutions are:
a=0, b=0 and a=2, b=2
from the 2nd equation:
LS = a/b + b/a
= (a^2 + b^2)/ab
= ( (a+b)^2 - 2ab)/ab
= (a^2 b^2 - 2ab)/ab
= ab(ab - 2)/ab
= ab - 2
---- and that is what you got, looks like we both are stuck at the same point.
another attempt:
from the first:
ab - b = a
b(a-1) = a
b = a/(a-1)
in 2nd:
LS = a/b + b/a
= a/((a/(a-1)) + (a/(a+1)(1/a)
= a(a-1)/a + 1/(a+1)
= a-1 + 1/(a+1)
= ((a-1)(a+1) + 1)/(a+1)
= (a^1 - 1 + 1)/(a+1)
= a^2/(a+1) <---- well that is interesting too
let's check a non-integer value
a non-integer solution would be a = 4/3, b=4
check: a+b = 4/3 + 4 = 16/3, and ab = 4(4/3) = 16/3 , OK in 1st equation
check in a/b + b/a = -1
LS = (4/3)/4 + 4/(4/3)
= 1/3 + 16/3 ≠ -1
All I needed was ONE case where the statement was false, I did
so there is something bogus with this question.