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alebra problem

by kat187 » Fri Mar 19, 2010 2:08 pm
hi I got a simple question for all you gurus out there.

if xy=1

then

2 ^(x+y)^2 / 2 ^(x-y)^2 =

for this can I substitute x=2 and y=1/2 ?
and rewrite it as


2^(x+y) (x+y)/ (x-y)(x-y)
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by papgust » Fri Mar 19, 2010 5:37 pm
No you cannot do that IMO.

Here's how you must do it.

if xy=1, then 2 ^(x+y)^2 / 2 ^(x-y)^2 = ?
(x+y)^2 = x^2 + y^2 + 2xy
(x-y)^2 = x^2 + y^2 - 2xy
Numerator --> 2^(x+y)^2. Whenever you face an exponent, Always start simplifying from the topmost exponent.
2^(x+y)^2 = 2^(x^2+y^2+2xy)
xy=1, 2^(x^2+y^2+2*1) = 2^(x^2+y^2+2) = 2^(x^2+y^2) * 2^2 = 2^(x^2+y^2) * 4

Denominator --> 2^(x-y)^2 = 2^(x^2+y^2-2xy) = 2^(x^2+y^2 - 2*1) = 2^(x^2+y^2-2) = 2^(x^2+y^2) * 2^(-2)


Numerator / Denominator = 2^(x^2+y^2) * 4 / 2^(x^2+y^2) * 2^(-2)

2^(x^2+y^2) gets cancelled out.

4 / 2^(-2) = [spoiler]4*4 = 16[/spoiler] IMO

Can you post the OA?
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by kat187 » Fri Mar 19, 2010 6:06 pm
=16
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by thephoenix » Fri Mar 19, 2010 9:46 pm
kat187 wrote:hi I got a simple question for all you gurus out there.

if xy=1

then

2 ^(x+y)^2 / 2 ^(x-y)^2 =

for this can I substitute x=2 and y=1/2 ?
and rewrite it as


2^(x+y) (x+y)/ (x-y)(x-y)
2 ^(x+y)^2 / 2 ^(x-y)^2=2 ^[(x+y)^2-(x-y)^2]=2^[(X+Y+X-Y)(X+Y-X+Y)]=2^[(2X)(2Y)]=2^4XY=2^4=16
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