For how many integers n is 2^n = n^2 ?
A. None
B. One
C. Two
D. Three
E. More than Three

\(2^n= n^2\) is true for 2 integers:
\(n=2\) --> \(2^2=2^2=4\);
\(n=4\) --> \(4^2=2^4=16\).

Well, \(2^2=2^2=4\) is obvious choices, then after trial and error you'll get \(4^2=2^4=16\) as well. But how do we know that there are no more such numbers? You can notice that when \(n\) is more than 4 then \(2^n\) is always more than \(n^2\) so \(n\) cannot be more than 4. \(n\) cannot be negative either as in this case \(2^n\) won't be an integer whereas \(n^2\) will be.

Answer: C.

NOTE: I think it's worth remembering that \(4^2=16=2^4\), I've seen several GMAT questions on number properties using this (another useful property \(8^2=4^3=2^6=64\)).

Hope it helps.
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=> 2^n = n^2
Taking nth root on both sides
=> 2 = (n^2)^1/n
=> 2 = n ^ 2/n
Lets consider positive even multiples of 2 for n (since LHS = 2)
For n = 2
=> 2 = 2 ^ 2/2 - First value that satisfier

For n = 4
=> 2 = 4 ^ 2/4 - Second value that satisfier

For n = 8
=> 2 = 8 ^ 2/8 - Doesnt satisfy

For n = 16
=> 2 = 16 ^ 2/16 - Doesnt satisfy

Two values. Ans = C
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\(n^2 = 2^n\)

So, \(n\) = \(2^\frac{n}{2}\)

Now, n must be Even and a multiple of 2

Plug in the even values only 2 and 4 remains...

\(2\) = \(2^\frac{2}{2}\)

\(4\) = \(2^\frac{4}{2}\)

Hence, answer will be (C) 2
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If n = 0:

2^n = 2^0 = 1 (recall that any nonzero number to the power of 0, is 1).
n^2 = 0^2 = 0.
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akash7gupta11

\(2^n = n^2 \)
What all does this tell you
(a) n cannot be negative as LHS 2^n will become fraction, while n^2 will remain integer.
(b) Since we have LHS as power of 2, the RHS or n will also be in terms of 2

So, let \(n=2^x\)....
\(2^{2^x}=(2^x)^2=2^{2x}.......2^x=2x.....2^{x-1}=x\)
Now moment x>2, the equality fails, so x=1 and 2

C
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