"Prove that if a^x =b^y =(ab)^xy , then x+y=1"
Taking logs of both sides gives:
xloga=ylogb=xylogab=xyloga+xylogb
Where do I go from here?
Thanks

We start with

(1) a^x=b^y

(2) a^x=(a b)^{x y}

We rewrite (2):

a^x=a^{x y}(b^y)^x

Substitute (1) into the right hand side of (2) to obtain

a^x=a^{x y}(a^x)^x

or

a^x=a^{x y+x2}

Assuming a is non zero (and so b) we can divide both sides by a^x to obtain

1=a^{x y+x2-x}

Assuming a is a real number and a > 0 then providing a is not 1 we must have

x y+x² - x=

and providing x is non-zero the result x+y=1 follows.

Thanks, the question was taken out of the P2 Chapter on logarithms, and was cited as being a previous exam question... so it should really have been correct! (perhaps it was specified correctly in the exam).

well you can solve this one by logarithm as well...

a^x =b^y
xloga=ylogb.........(1)

a^x =(ab)^xy
xloga=xyloga+xylogb....(2)

from(1) and (2),
xloga=xyloga+x² loga
(x² +xy-x)loga=0
therefore,
x² +xy-x=0

giving you x+y=1

as you see this is just a modification of William's proof...

love arun

Ahh... yeah, I didn't see the substitution of x^2loga for xylogb. Thanks again for both answers