Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 24 May 2013, 14:33

# A very difficult math

Author Message
TAGS:
Intern
Joined: 21 Aug 2011
Posts: 32
Followers: 0

Kudos [?]: 1 [0], given: 0

A very difficult math [#permalink]  23 Aug 2011, 18:09
00:00

Question Stats:

50% (02:08) correct 50% (00:48) wrong based on 1 sessions
If x and K are integers and (12^x)(4^2x+1)=(2^k)(3^2), what is the value of K?

a) 5
b)7
c)10
d)12
e)14

Pls explain me and there gotta be a simple way to solve this?
Manager
Joined: 25 Dec 2010
Posts: 88
Followers: 0

Kudos [?]: 9 [0], given: 2

Re: A very difficult math [#permalink]  23 Aug 2011, 18:15
considering 4 ^2x+1 as 4^(2x+1), E it is

(write 12^x as 2^2x. 3^x)

the eqn will drill down to

3^x . 2^(6x+2) = 2^k.3^2
hence x=2
and 6x+2=k
so k = 14
Manager
Status: On...
Joined: 16 Jan 2011
Posts: 193
Followers: 2

Kudos [?]: 26 [0], given: 62

Re: A very difficult math [#permalink]  23 Aug 2011, 18:57
Jasonammex wrote:
If x and K are integers and (12^x)(4^2x+1)=(2^k)(3^2), what is the value of K?
(12^x)(4^(2x+1))=(2^k)(3^2)
==> (3^x)(4^x)(4^(2x+1))=(2^k)(3^2)
==> (3^x)(4^(3x+1))=(3^2)(4^(k/2))

So 3^x = 3^2 ==>x = 2
4^(3x+1) = 4^(k/2) ==> k = 2(3x+1) = 2(3*2+1) = 14
So OA -E
a) 5
b)7
c)10
d)12
e)14

Pls explain me and there gotta be a simple way to solve this?

Hope the above solution helps.
_________________

Labor cost for typing this post >= Labor cost for pushing the Kudos Button
kudos-what-are-they-and-why-we-have-them-94812.html

Intern
Joined: 02 Aug 2011
Posts: 11
Followers: 0

Kudos [?]: 1 [0], given: 1

Re: A very difficult math [#permalink]  23 Aug 2011, 18:59
(12^x)(4^(2x+1)) = (2^k)(3^2)

we want to get the bases to match, so we can then solve for k in the exponent.

12 can be broken down as 3 x 2 x 2. As the rules for exponents go, if the exponents are the same then you can perform the operation as normal with the bases so 12^x = (2^x)(2^x)(3^x).
you also want the 4 down to a base of 2, and square root of 4 is 2 so 2^2=4, so again using another exponent rule : (2^2)^(2x+1) we multiply the exponents, and get 2^(4x+2)

(2^2x)(3^x)(2^(4x+2)) = (2^k)(3^2)

Now we've got two terms with a base of 2, so we multiply them : which means we add the exponents:

(2^(2x+4x+2))(3^x) = (2^k)(3^2)

Now we can compare both sides, since the bases are the same.
So we can assume, since 3^x = 3^2, then x=2 and 2^(2x+4x+2) = 2^k, therefore 2x+4x+2 = k

Substituting 2 for x, we get

2(2) + 4(2) + 2 = k
4 + 8 + 2 = k
14 = k

Last edited by meshell on 23 Aug 2011, 19:55, edited 1 time in total.
Director
Joined: 01 Feb 2011
Posts: 791
Followers: 11

Kudos [?]: 63 [0], given: 42

Re: A very difficult math [#permalink]  23 Aug 2011, 19:01
12^x * 4^(2x+1) = 2^k * 3^2

3^x 2^2x 2^(4x+2) = 2^k * 3^2

equating powers of 2 and 3 on both sides we have

6x+2=k
x=2

=> k =14
Intern
Joined: 21 Aug 2011
Posts: 32
Followers: 0

Kudos [?]: 1 [0], given: 0

Re: A very difficult math [#permalink]  23 Aug 2011, 20:07
Thank you you guys, I just don't know 12^x could be divided into 3^x(4^x).
Re: A very difficult math   [#permalink] 23 Aug 2011, 20:07
Similar topics Replies Last post
Similar
Topics:
Is Kaplan math really that difficult? 0 03 Aug 2003, 15:52
2 Combinatorics (very difficult) 4 08 Sep 2008, 09:39
Difficult Math Question 2 01 Oct 2008, 18:37
A difficult math, pls help 2 24 Jul 2011, 03:48
A very very difficult math 3 24 Aug 2011, 02:02
Display posts from previous: Sort by