if a^x= b^y = (ab)^(xy) prove that x+y =1

ln(a^x) = ln(b^y) = ln((ab)^(xy))
xln(a) = xyln(ab)
ln(a) = yln(ab) = y(ln(a) + ln(b))
y = ln(a)/(ln(a)+ln(b))
with same analysis for ln(b^y):
ln(b) = x(ln(a) + ln(b))x = ln(b)/(ln(a)+ln(b))
x + y = (ln(a) + ln(b))/(ln(a) + ln(b)) = 1