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

Question

A. None 
B. One 
C. Two 
D. Three 
E. More than Three

Bunuel · Accepted Answer

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.

Abhishek009 · Answer

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

shrouded1 · Answer

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.

https://img243.imageshack.us/img243/3857/graph1.gif

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

https://img713.imageshack.us/img713/3756/graph2.gif

ravsg · Answer

=> 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

AmritaSarkar89 · Answer

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.

Bunuel · Answer

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.

BoundMan · Answer

Hello  chetan2u

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

chetan2u · Answer

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

CrackverbalGMAT · Answer

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} 
eq 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

MathRevolution · Answer

2^n = n ^2  for n = 2 and 4

=>  2^2 = 2^2 = 4

=>  2^4 = 4^2 = 16

Answer C

Basshead · Answer

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.

BrushMyQuant · Answer

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

https://www.youtube.com/watch?v=9la4tJI7ZiI

sachi-in · Answer

This is way simple if we simply draw the grap for 2^n and n^2 Consider the X axis as N then for: n > 0 : n^2 will catchup with 2^n at n = 2 ( intersection point and never ever meet after that point, imagine 10^10 is much higher than 2^10 ) n < 0 : n^2 will be symmetric ( from 0 it will be ever increasing ) but 2^n is ever decreasing ( from 1 it decreases to zero ) so they will meet only once and then never ever meet They meet twice once for n > 0 and another time for n < 0

Aboyhasnoname · Answer

Let's keep it simple....

Since n can't be negative... Start with 0 
2^0 = 0^2 ... Nope 
2^1 = 1^2...Nope 
2^2 = 2^2...Yes!....I
2^3=3^2..Nope... 
2^4 = 4^2 yes..... II 
..The rest will not be equal... 
So C!

AryakiSekera · Answer

Would 0 not be considered and integer for this question?

Bunuel · Answer

0 is indeed an integer. However, it does  not  satisfy the given equation:

2^0 = 1, while 0^2 = 0, so the two sides are not equal.

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

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