Hi All,

Please help in the following exponents question.

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

(A) None

(B) One

(C) Two

(D) Three

(E) More than three

Answer is C

Thanks & Regards

Sachin

## Exponents

##### This topic has expert replies

- sachin_yadav
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- GMATGuruNY
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Since the left side of the equation has a base of 2, the right side also needs a base of 2, with each side raised to the same power.sachin_yadav wrote:Hi All,

Please help in the following exponents question.

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

(A) None

(B) One

(C) Two

(D) Three

(E) More than three

Answer is C

Thanks & Regards

Sachin

n=2:

2^n = 2^2

n^2 = 2^2

This works.

n=2^2:

2^(2^2) = 2^4

(2^2)^2 = 2^4

This works.

n=2^3:

2^(2^3) = 2^8

(2^3)^2 = 2^6

Doesn't work.

n=2^4:

2^(2^4) = 2^16

(2^4)^2 = 2^8.

Doesn't work.

Now we can see that as the exponent increases, the difference between 2^n and n^2 only becomes greater.

Thus, for only 2 values does 2^n = n^2:

n=2 and n= 2^2 = 4.

The correct answer is C.

Last edited by GMATGuruNY on Wed Mar 30, 2011 11:50 am, edited 3 times in total.

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- 6983manish
- Senior | Next Rank: 100 Posts
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By plugging in ,sachin_yadav wrote:Hi All,

Please help in the following exponents question.

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

(A) None

(B) One

(C) Two

(D) Three

(E) More than three

Answer is C

Thanks & Regards

Sachin

for n = 1 , 2^1 = 2 <> 1^2 = 1

for n = 2 , 2^2 = 4 = 2^2 = 4

for n = 3 , 2^3 = 8 <> 3^2 = 9

for n = 4 , 2^4 = 16 = 4^2 = 16

for n = 5 , 2^5 = 32 <> 5^2 = 25

Now onwards it will start differing by bigger difference.

Hence , its true for 2 and 4.

Answer shouls be C.

- manpsingh87
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taking square root on both sides we have; 2^n/2=n;sachin_yadav wrote:Hi All,

Please help in the following exponents question.

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

(A) None

(B) One

(C) Two

(D) Three

(E) More than three

Answer is C

Thanks & Regards

Sachin

now consider n=2; 2^2/2=2; hence n=2 yields solution.

now consider n=4; 2^4/2=4; hence n=4 yields solution.

for any other value of n, we won't get any desired solution.!!

hence C

O Excellence... my search for you is on... you can be far.. but not beyond my reach!

- sachin_yadav
- Master | Next Rank: 500 Posts
**Posts:**212**Joined:**06 Dec 2010**Location:**India**Thanked**: 5 times**Followed by:**1 members

I also solved the question with the substitution method and got the answer but did not find logical.GMATGuruNY wrote:

Since the left side of the equation has a base of 2, the right side also needs a base of 2, with each side raised to the same power.

By plugging in ,

I think Mitch's explanation is quite logical because this is what i wanted to know. Thanks Mitch, really appreciate for your help.for n = 1 , 2^1 = 2 <> 1^2 = 1

for n = 2 , 2^2 = 4 = 2^2 = 4

for n = 3 , 2^3 = 8 <> 3^2 = 9

for n = 4 , 2^4 = 16 = 4^2 = 16

for n = 5 , 2^5 = 32 <> 5^2 = 25

Regards

Sachin

### GMAT/MBA Expert

- lunarpower
- GMAT Instructor
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this problem can also be solved very easily by brute force: just plug a bunch of integers into it till you start seeing relevant patterns.

first, it should become clear rather quickly that negative numbers can't solve this equation; if n is a negative number, then 2^n will be a fraction, but n^2 will be an integer.

so try 0, 1, 2, 3, ...

if you try these numbers, you'll notice that you get correct solutions for n = 2 and n = 4. after that point, you will notice that 2^n starts to grow MUCH more rapidly than n^2 -- a pattern that continues forever.

--

the primary purpose of this problem is to spit in the faces of students whose

there are in fact tons of these problems. for a 1.5-hour study hall lecture on problems for which algebra is ineffective, watch the NOVEMBER 4, 2010 lecture at the following website:

https://www.manhattangmat.com/thursdays-with-ron.cfm

first, it should become clear rather quickly that negative numbers can't solve this equation; if n is a negative number, then 2^n will be a fraction, but n^2 will be an integer.

so try 0, 1, 2, 3, ...

if you try these numbers, you'll notice that you get correct solutions for n = 2 and n = 4. after that point, you will notice that 2^n starts to grow MUCH more rapidly than n^2 -- a pattern that continues forever.

--

the primary purpose of this problem is to spit in the faces of students whose

*only*approach to equations is algebra.there are in fact tons of these problems. for a 1.5-hour study hall lecture on problems for which algebra is ineffective, watch the NOVEMBER 4, 2010 lecture at the following website:

https://www.manhattangmat.com/thursdays-with-ron.cfm

Ron has been teaching various standardized tests for 20 years.

--

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Pueden hacerle preguntas a Ron en castellano

Potete chiedere domande a Ron in italiano

On peut poser des questions Ã Ron en franÃ§ais

Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

--

*Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.*Yves Saint-Laurent

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