The expression x[n]y is defined for positive values of x and

Question

The expression x y is defined for positive values of x and y and for positive integer values of n as follows:

x y = x^y
If n is odd, x y = (x y)^x
If n is even, x y = (x y)^y

If y = ½ and x y = 2, then x =

A. ¼
B. ½
C. 1
D. 2
E. 4

Bunuel · Accepted Answer

Given:

x y = x^y

If  n  is even,  x y = (x y)^x  ( notice that it's the same as: if n is odd, x y = (x y)^x );

If  n  is odd,  x y = (x y)^y  ( notice that it's the same as: if n is even, x y = (x y)^y )

The question asks: If y = ½ and x y = 2, then x =

Since 4 is even then  x y=(x y)^x ;

Since 3 is odd then  (x y)^x=((x y)^y)^x=(x y)^{xy} ;

Since 2 is even then  (x y)^{xy}=((x y)^x)^{xy}=(x y)^{x^2y} ;

Since  x y = x^y  then  (x y)^{x^2y}=(x^y)^{x^2y}=x^{x^2y^2} .

So, finally we have that:  x y=x^{x^2y^2}=2 . Now, as  y=\frac{1}{2}  then substituting this value into  x^{x^2y^2}=2  gives  x^{\frac{x^2}{4}}=2 .

Take to the fourth power:

x^{x^2}=16 ;

x=2  or  x=-2 .

Answer: D.

GyanOne · Answer

x y = (x y)^y = ((x y)^x)^y = ((((x y)^y)^x)^y) = (x y)^(x*y^2) = (x y)^x

Now x y = x^(1/2)

=> (x^(1/2))^x = 2
=> x^(x/2) = 2
=> x^x = 4 
=> x = 2

Option D

saleem1992 · Answer

Bunuel, how can x be = -2 ?

(x)^(x)^(2)  ---->  -2^(-2)^2  = -1/16 .. Am I right?

Bunuel · Answer

No.

(-2)^(-2)^2 = (-2)^4 = 16.

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:

a^{m^n}=a^{(m^n)}  and not  {(a^m)}^n , which on the other hand equals to  a^{mn} .

NaeemHasan · Answer

Hi, Bunuel!
I am unable to understand the colored line. How does power shift from Y to X in the first line and X to Y in the second line?

Bunuel · Answer

This is given in the definition of the function: x y is defined for positive values of x and y and for positive integer values of n as follows:
x y = x^y
If n is odd, x y = (x y)^ x 
If n is even, x y = (x y)^ y

NaeemHasan · Answer

Bunuel ,
But you have converted x y = (x y)^x to x y = (x y)^y when n is odd and same when n is even.

Bunuel · Answer

Where did I do that? Maybe you are confused by double exponents?

Since 4 is even then x y=(x y)^ x ;

Since 3 is odd then (x y)^x=((x y)^ y )^x=(x y)^(xy);

Since 2 is even then (x y)^(xy)=((x y)^ x )^(xy)=(x y)^(x^2y);

KarishmaB · Answer

x y = (x y)^x = (x y)^{xy} = (x y)^{x^2y} = (x^y)^{x^2y} = x^{(xy)^2}

Since y = 1/2, we get  2 = x^{(x/2)^2}

Look for a value of x from the options which will satisfy this equation. Since the power has x/2 but LHS is an integer, x should be a multiple of 2.

Try x = 2, you get  2 = 2^{(2/2)^2} = 2^1 
Satisfies

Answer (D)

SiffyB · Answer

Hi  VeritasKarishma ,

Since 4 is even, shouldn't  x y = (x y)^y   ??

Kindly help.

KarishmaB · Answer

SiffyB

Note again how the operation is defined.

If n is odd, x   y = (x y)^x
If n is even, x   y = (x y)^y

Yes, 4 is even but 4 is  , not n

You are given that x y = 2
This means x y = 2
So n+1 = 4. But then, n = 3 (odd)

Hence x y = (x y)^x

HoudaSR · Answer

Hi  Bunuel

Can you please explain to me why x can be equal to -2? While solving it appeared to me that x=2 and I couldn't see the case where it can actually equal -2.

Thank you in advance for your valuable support.

Bunuel · Answer

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
 
 a^{m^n}=a^{(m^n)}  and not  {(a^m)}^n , which on the other hand equals to  a^{mn} .
 
So:
 
 (a^m)^n=a^{mn} 
 
 a^{m^n}=a^{(m^n)}  and not  {(a^m)}^n .
 
 According to the above:

x^{x^2}=x^{(x^2)} . Now, if x = -2, we'd get:

(-2)^{(-2)^2} = (-2)^{((-2)^2)}= (-2)^4 = 16.

HoudaSR · Answer

If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
 
 a^{m^n}=a^{(m^n)}  and not  {(a^m)}^n , which on the other hand equals to  a^{mn} .
 
So:
 
 (a^m)^n=a^{mn} 
 
 a^{m^n}=a^{(m^n)}  and not  {(a^m)}^n .
 
 According to the above:

x^{x^2}=x^{(x^2)} . Now, if x = -2, we'd get:

(-2)^{(-2)^2} = (-2)^{((-2)^2)}= (-2)^4 = 16.

Thank you  Bunuel  for your feedback.

Following your explaination, I could see that -2 is also a solution to x but while solving this problem it wouldn't have been obvious to me unless I would have questioned if it is.

From the way I solved it in the picture below, I couldn't see at first that -2 can be a solution! Any feedback/tips?

Bunuel · Answer

Since x = 2 is a solution of  x^{x^2}=16 , then you could guess that x = -2 must also satisfy the equation because x there is in even power.

HoudaSR · Answer

Probably the reason why it wasn't obvious to me is because the unkown is on the exponent level. However, thinking of the even power as you suggested is helpful regardless of where that unkown sits. Thank you so much  Bunuel  for your support. It has been very much valuable.

onlymalapink · Answer

Given:

If is even, (notice that it's the same as: if n is odd, x y = (x y)^x);

If is odd, (notice that it's the same as: if n is even, x y = (x y)^y)

I'm not a sorcerer there is no way I would 9rganically ever understand this weird way of understanding a question I would never assume n is n plus 1

The expression x[n]y is defined for positive values of x and

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