If f(x) = x^x,then f(f(x)) =

Please note:

\(2^{3^2} = 2^{(3^2)} = 2^9 = 512\)

\((2^3)^2 = 2^{(3*2)} = 2^6 = 64\)

OR

\((2^3)^2 = 8^2 = 64\)

Power calculations go from top to bottom, if brackets are not provided

Thanks for your reply

Would you try this question by plugging in. The question looked good for plugging in options but I am not getting the answer.

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Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functions

hi Bunuel

the use of exponents is little bit trickier to me here

please let me understand...

f [ f(x) ] to the power f(x)
that is [ f (x) ] is raised to power f(x)..

NOT, f (x) is raised to power f(x), if this is so, then it may mean, in f(x), "x" is raised to the power f(x), and thus may end up with result looking like option A

So, the expression here is meant to be "[ f (x) ]" (the whole) is raised to the power f(x)" and thus option D is right
Am I right or wrong ....? please help me get rid of my doubts ...

thanks in advance, man

Paresh HERE explains it so perfectly that I have nothing to add.

Since f(x) = x^x, f(f(x)) = f(x^x) = (x^x)^(x^x) = x^(x * x^x) = x^(x^(x + 1)).

Alternate Solution:

Let’s take x = 2. Then, f(2) = 2^2 = 4 and f(f(2)) = f(4) = 4^4 = 2^8. Let’s test each answer choice to see which one(s) simplify to 2^8:

A) x^x^x^x

If x = 2, we get 2^2^2^2 = 2^2^4 = 2^16. A is not correct.

B) x^x^x

If x = 2, we get 2^2^2 = 2^4. B is not correct.

C) x^x^x^2

If x = 2, then we get 2^2^2^2 which is the same as answer choice A. C is not correct.

D) x^x^(x + 1)

If x = 2, then we get 2^2^(2 + 1) = 2^2^3 = 2^8. D might be correct.

E) (x^x)^x

If x = 2, we get (2^2)^2 = 4^2 = 2^4. E is not correct.

Since D is the only answer choice that yields 2^8 when x is 2, D must be the correct answer choice.

Answer: D

\(x^{x^{(x^x)}\)
\(=x^{x*x^x}\)
\(=x^{x^{x+1}}\)

IMO D

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Even though your explanation seems plausible, but I don't think we follow such rules in our algebra classes.

Common steps of f(f(x))
(1) Solve f(x)
(2) Put the value of (1) in f(f(x)) thus making it f((solutionOf1))

Thus f(x) = \(x^x\)
and
f(f(x)) = \(x^x^x^x\)

or let y = \(x^x\)
f(y) = \(y^y\)

Thoughts?

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