For how many integers n is 2^n = n^2 ?

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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It helps to know that the function 2^x is more expansive than x^2 for large positive x and converges quickly to 0 for negative x.



So we know we only have to check small values of x. For positive x, it is easy to see this is true for x=2,4 and then the function 2^x explodes

For negative x, 2^-1 is less than -1^2 already so no negative integers can satisfy the equality

Answer is (c) or 2


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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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This question might very well baffle us under the exam stress . So what is the methodology.
We can assume that the number wont be very large as - the bigger the numbers will get the difference between the two algebraic expression will increase.
STart with n=0
we realise 2^0 is not equal to 0^2.
The continue with 1, 2, 3 4, 5, 6, 7, 8 by then you will get n=2, 4 suits the criterion, others don't and any bigger number will go super off limit.

trust me it took my 59 sec to do it using this long method.

\(n^2 = 2^n\)

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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Hello chetan2u

Is there another approach to solving this question? I do not want to rely on hit and trial method.

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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\(2^n = n^2\) therefore, \(\frac{2^n}{n^2} = 1\)

Odd powers are out as the denominator has to be even.

When n = 2, \(\frac{2^2}{2^2} = \frac{4}{4} = 1\)

When n = 4, \(\frac{4^2}{2^4} = \frac{16}{16} = 1\)

When n = 6, \(\frac{2^6}{6^2} = \frac{64}{36} \neq 1\)

Our search stops here as for all other multiples, the difference between the numerator and denominator will keep increasing.


Therefore only 2 possible values


Option C

Arun Kumar
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\(2^n = n ^2\) for n = 2 and 4

=> \(2^2 = 2^2 = 4\)

=> \(2^4 = 4^2 = 16\)

Answer C
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For how many integers n is 2^n = n^2 ?

We can plug in numbers starting with 0:

2^0 = 1
0^2 = 0

2^2 = 4
2^2 = 4

2^3 = 8
3^2 = 9

2^4 = 16
4^2 = 16

There are two integers for which 2^n = n^2.
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\(2^n\) = \(n^2\) for two values of n

n = 2
\(2^2\) = \(2^2\) = 4 and

and n = 4
\(2^4\) = \(4^2\) = 16

So, Answer will be C
Hope it helps!

Watch the following video to learn the Basics of Exponents


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