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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]

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23 Aug 2011, 18:09
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77% (01:45) correct 23% (01:28) wrong based on 31 sessions

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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]

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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
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Re: A very difficult math [#permalink]

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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.
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Re: A very difficult math [#permalink]

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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.
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Re: A very difficult math [#permalink]

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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
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Re: A very difficult math [#permalink]

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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
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