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If x and k are integers and 12^x*4^(2x+1)=2^k*3^2, what is

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If x and k are integers and 12^x*4^(2x+1)=2^k*3^2, what is [#permalink] New post 23 Aug 2011, 17:09
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A
B
C
D
E

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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
[Reveal] Spoiler: OA
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Re: A very difficult math [#permalink] New post 23 Aug 2011, 17:15
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
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Re: A very difficult math [#permalink] New post 23 Aug 2011, 17:57
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.
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Re: A very difficult math [#permalink] New post 23 Aug 2011, 17:59
(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

Answer : E

Last edited by meshell on 23 Aug 2011, 18:55, edited 1 time in total.
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Re: A very difficult math [#permalink] New post 23 Aug 2011, 18:01
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
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Re: A very difficult math [#permalink] New post 23 Aug 2011, 19:07
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, 19:07
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