Prove that if a^x = b^y = (ab)^xy, then x + y = 1?

Question

Prove that if a^x = b^y = (ab)^xy, then x + y = 1

Answer 1

a^x = (ab)^(xy)
lg(a^x) = lg((ab)^(xy))
x.lg(a)=xy.lg(ab)
y=lg(a)/lg(ab)
b^y = (ab)^(xy)
lg(b^y) = lg((ab)^(xy))
y.lg(b) = xy.lg(ab)
x = lg(b)/lg(ab)
x+y = lg(b)/lg(ab)+lg(a)/lg(ab)
x+y = [lg(b)+lg(a)]/lg(ab)
but: lg(a.b) = lg(a) + lg(b)
x+y = (lg(a) + lg(b))/(lg(a) + lg(b))=1

Answer 2

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Answer 3

xloga =ylogb = xy log ab = xy loga + xylogb
loga/y = logb/x = log ab
x = logb/logab
y = loga/logab
x+y = (logb +loga)/logab=logab/logab = 1

Answer 4

a^x = (ab)^xy
ln[a^x] = ln[(ab)^xy]
x ln[a] = xy{ln[a] + ln[b]}
y = ln[a]/{ln[a] + ln[b]}

b^y = (ab)^xy
ln[b^y] = ln[(ab)^xy] = xy{ln[a] + ln[b]} = y ln[b]
x = ln[b]/{ln[a] + ln[b]}

x+y = {ln[a]+ln[b]}/{ln[a]+ln[b]} = 1
because they have a common denominator