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Exponents

by sachin_yadav » Wed Mar 30, 2011 10:46 am
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

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by GMATGuruNY » Wed Mar 30, 2011 10:55 am
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
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.

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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by 6983manish » Wed Mar 30, 2011 11:00 am
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
By plugging in ,

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.

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by manpsingh87 » Wed Mar 30, 2011 9:36 pm
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
taking square root on both sides we have; 2^n/2=n;

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
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by force5 » Wed Mar 30, 2011 11:55 pm
i think thats the best way...

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by sachin_yadav » Thu Mar 31, 2011 6:49 am
Thank you very much for your replies and really appreciate for your help. From my point of view the easy and the logical method would be to make both the bases same.
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.
I also solved the question with the substitution method and got the answer but did not find logical.

By plugging in ,
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
I think Mitch's explanation is quite logical because this is what i wanted to know. Thanks Mitch, really appreciate for your help.

Regards
Sachin

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by lunarpower » Fri Apr 01, 2011 4:06 am
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.

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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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