Answer:
Below.
Explanation:
Prove that if
a^x = b^y = (ab)^(xy) then x + y = 1.
From the second equation:
y = 1 - x
So we have:
a^x = b^y = b^(1-x)
Taking logs:
x ln a = (1-x)ln b
x ln a + x ln b = ln b
x = ln b / ( ln a + ln b)
x = ln b / ln ab.
This is true if x + y = 1.
Now consider b^y = (ab)^(xy):
b^(1 - x) = (ab)^(x(1- x)
ln b - x ln b = x(1 - x)ln ab
ln b - x ln b = x ln ab - x^2 ln ab
Now we substitute for x = ln b / ln ab in this expression.
If left side = right side then we have proved this identity and therefore the original one.
Left side = ln b - (ln b / ln ab) * ln b = ln b - (ln b)^2 / ln ab.
Right side = (ln b / ln ab) * ln ab - [(ln b)^2 / (ln ab)^2] * ln ab
= ln b - (ln b)^2 / ln ab.
Left side = right side so this identity is true, and so the original one
( x + y = 1) is true also.