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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Exponents
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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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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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6983manish
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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.
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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
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sachin_yadav
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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.
By plugging in ,
Regards
Sachin
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
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lunarpower
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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 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
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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Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
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Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions à Ron en français
Voit esittää kysymyksiä Ron:lle myös suomeksi
--
Quand on se sent bien dans un vêtement, tout peut arriver. Un bon vêtement, c'est un passeport pour le bonheur.
Yves Saint-Laurent
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Learn more about ron
