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ax=by=(ab)xy
[Editor: Throghout this question take a,b > 0]
By Philip Ellison on Monday, January 07, 2002 - 04:57 pm:

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

By William Astle on Monday, January 07, 2002 - 06:08 pm:

The deduction can't necessarily be made. If a=b=1 for example then x and y can be totally independent.

We start with

(1) ax=by
(2) ax=(ab)xy

We rewrite (2):

ax=axy(by)x

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

ax=axy(ax)x

or

ax=axy+x2

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

1=axy+x2-x

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

xy+x2-x=0

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

By Philip Ellison on Monday, January 07, 2002 - 07:31 pm:

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).

By Arun Iyer on Monday, January 07, 2002 - 07:38 pm:

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

ax=by
xloga=ylogb.........(1)

ax=(ab)xy
xloga=xyloga+xylogb....(2)

from(1) and (2),
xloga=xyloga+x2loga
(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

By Philip Ellison on Monday, January 07, 2002 - 08:54 pm:

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