Exponents

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sachin_yadav: Master | Next Rank: 500 Posts; Posts: 212; Joined: Mon Dec 06, 2010 12:52 amFollowed by: 1 member; Location: India; Thanked: 5 times

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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Source: Beat The GMAT — Problem Solving | Wed Mar 30, 2011 10:46 am

GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

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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6983manish: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Mon Mar 07, 2011 12:27 am; Thanked: 6 times

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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manpsingh87: Master | Next Rank: 500 Posts; Posts: 436; Joined: Tue Feb 08, 2011 3:07 amFollowed by: 6 members; Thanked: 72 times

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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force5: Legendary Member; Posts: 582; Joined: Tue Mar 08, 2011 12:48 amFollowed by: 6 members; Thanked: 61 times; GMAT Score:740

by force5 » Wed Mar 30, 2011 11:55 pm

i think thats the best way...

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sachin_yadav: Master | Next Rank: 500 Posts; Posts: 212; Joined: Mon Dec 06, 2010 12:52 amFollowed by: 1 member; Location: India; Thanked: 5 times

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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lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

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

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