a^x =b^y =(ab)^xy

[Editor: Throughout this question take a,b > 0]

By Philip Ellison on Monday, January 07, 2002 - 04:57 pm:

"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

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

(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

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

$a^{x} = a^{x y + x^{2}}$

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

$1 = a^{x y + x^{2} - x}$

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

$x y + x^{2} - x =$

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...

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

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