Niece teaser on Exponents

Question

Got me stumped for no good reason.

BarneyStinson · Answer

2^4 = 16 is the answer.

Basically (2^x)/(2^y) = 2^(x-y) and then, there's some algebra which I am sure, you can figure it out yourselves.

mrblack · Answer

x and y both equal to 1 or -1.

Plug the numbers in and you end up with 2^4=16 in both cases.

DestinyChild · Answer

yup! I had... just put it out there...!

dmetla · Answer

if u solve it its 2^4=16

cameraknox · Answer

I still dont get. Can you please fully explain...thanks.

vladik210 · Answer

Sure.

Here is what you need to know:
A.  (x+y)^2 = x^2 +2xy +y^2 
B.  (x-y)^2 = x^2 -2xy +y^2 
C. As BarneyStinson has mentioned you need to know that  (2^x)/(2^y) = 2^{x-y}

Now solve the problem using the formulas provided above.

2^{x^2 +2xy +y^2}/2^{x^2 -2xy +y^2} ;
 2^{x^2 +2*1 +y^2}/2^{x^2 -2*1 +y^2} ; (Remember xy = 1)

By using formula C you get  2^{x^2 +2 +y^2 -x^2 -(-2) -y^2} 
And finally after canceling  x^2  and  y^2  out, you get  2^{2-(-2)} 
What in turn is equal to  2^4 or just 16

I hope this explanation helps you.

johnhillescobar · Answer

Thanks DestinyChild. I enjoyed a lot this problem.

Since \(\frac{2^{(x+y)^2}} {2^{(x-y)^2}}\) then, you can get two quadratic equations from the exponents: \({(x+y)^2 = x^2 + 2xy + y^2}\) and \({(x-y)^2 = x^2 - 2xy + y^2}\)

Therefore, \(X^2\) and \(y^2\) both in the numerator and denominator should be canceled, which is equivalent to \(\frac{2^{2xy}} {2^{-2xy}}\). This implies that \(\frac{2^{2xy}} {\frac{1} {2^{2xy}}}\).

Because \(xy = 1\) you can calculate the exponents values to get \(\frac{2^2} {\frac{1} {2^2}}\) or \(\frac{4} {\frac{1} {4}}\). Finally, applying the double C you'll get 16.

Hope this helps

DestinyChild · Answer

glad you liked it... i bet you can still get this wrong on any day... have an argument with your gf\wife and try it...

or try it after shoveling 28" of snow...

johnhillescobar · Answer

Of course DestinyChild...it's possible!!!! :-D

shrivastavarohit · Answer

Pretty late however this is simpler, please make a lots of noise in case approach is not correct.

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

Represent this eq. in non fraction format since base is 2 (same) the eq. should look as below.

xy = 1;

=2^(x^2 + y^2 + 2xy) - (x^2+y^2-2xy)
=2^4xy

Since xy=1

=2^4 = 16

nickk · Answer

Not necessarily, the question doesn't state that they are both integers.

dimitri92 · Answer

it doesn't matter if they are integers or not ....the final answer doesn't change

Niece teaser on Exponents

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