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

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

Since 4 is even then x[4]y=(x[3]y)^x;

Since 3 is odd then (x[3]y)^x=((x[2]y)^y)^x=(x[2]y)^(xy);

Since 2 is even then (x[2]y)^(xy)=((x[1]y)^x)^(xy)=(x[1]y)^(x^2y);

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:

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

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.

x[4]y = (x[3]y)^y = ((x[2]y)^x)^y = ((((x[1]y)^y)^x)^y) = (x[1]y)^(x*y^2) = (x[1]y)^x

Now x[1]y = x^(1/2)

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

Option D

Bunuel, how can x be = -2 ?

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

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 \(a^{mn}\).

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?

This is given in the definition of the function: x[n]y is defined for positive values of x and y and for positive integer values of n as follows:
x[1]y = x^y
If n is odd, x[n+1]y = (x[n]y)^x
If n is even, x[n+1]y = (x[n]y)^y

Bunuel,
But you have converted x[n+1]y = (x[n]y)^x to x[n]y = (x[n-1]y)^y when n is odd and same when n is even.

Hi VeritasKarishma,

Since 4 is even, shouldn't \(x[4]y = (x[3]y)^y \) ??

Kindly help.

SiffyB

Note again how the operation is defined.

If n is odd, x[n+1]y = (x[n]y)^x
If n is even, x[n+1]y = (x[n]y)^y

Yes, 4 is even but 4 is [n+1], not n

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

Hence x[4]y = (x[n]y)^x

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.

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:

\((-2)^{(-2)^2} = (-2)^{((-2)^2)}= (-2)^4 = 16.\)[/quote]


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?

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.