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

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If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 11 Mar 2014, 02:57
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E

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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 11 Mar 2014, 02:58
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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 11 Mar 2014, 20:11
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Answer = (E) 14

Re-write as

\(3^x . 4^x . 4^{(2x+1)} = 2^k. 3^2\)

LHS & RHS has only 1 base of 3, so x = 2

\(4^2 . 4^5 = 2^k\)

\(4^7 = 2^k\)

K = 14
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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 11 Mar 2014, 03:06
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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


Sol: The given expression can be reduced to (3^x*4^x)4^(2x+1)= 2^k*3^2

So we see that 3^x=3^2 or x=2

Also 4^x*4^(2x+1) can be written as 4^3x+1 =2^k or 2^2(3x+1)= 2^k or k=2*(3x+1) or k=14

Ans is E

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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 11 Mar 2014, 07:48
1
Its E..

12=3*2^2

4^2x+1=2^2(2x+1)

well will get x=2..

and 2x+4x+2=k

6x+2=k
x=2
k=14
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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 12 Feb 2018, 17:37
Bunuel wrote:
The Official Guide For GMAT® Quantitative Review, 2ND Edition

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


Let’s simplify the given equation:

(3 * 2^2)^x * (2^2)^(2x + 1) = 2^k * 3^2

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

3^x * 2^(6x + 2) = 3^2 * 2^k

Equating the exponents with the like bases, we see that:

x = 2 and 6x + 2 = k

Thus, k = 6(2) + 2 = 14.

Answer: E
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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 19 Feb 2018, 11:40
Bunuel wrote:
SOLUTION

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

\((12^x)(4^{2x+ 1})= (2^k)(3^2)\);

\((2^{2x}*3^x)(2^{2(2x+ 1)})= (2^k)(3^2)\);

\((2^{2x+2(2x+ 1)}*3^x)= (2^k)(3^2)\);

Equate the powers of 3 --> \(x=2\);
Equate the powers of 2 --> \(2x+2(2x+ 1)=k\) --> \(6x+2=k\) --> \(k=14\).

Answer: E.



Hi Bunuel

\((2^{2x+2(2x+ 1)}*3^x)= (2^k)(3^2)\); after this

\((2^{6x+2}*3^x)= (2^k)(3^2)\); <--- i get this

how do you Equate the powers of 3 --> \(x=2\); :? where do you see power of 3 ? :? how you get 2 :?

how do you Equate the powers of 2 --> \(2x+2(2x+ 1)=k\) --> \(6x+2=k\) --> \(k=14\). and get 14 :?

i know this rule -->

if \(a^x = a^y\) then \(x = y\)

in other words if the bases match then exponents are equal.

please explain :) \(thank you ^ {1000}\)

:)
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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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New post 19 Feb 2018, 11:48
1
dave13 wrote:
Bunuel wrote:
SOLUTION

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

\((12^x)(4^{2x+ 1})= (2^k)(3^2)\);

\((2^{2x}*3^x)(2^{2(2x+ 1)})= (2^k)(3^2)\);

\((2^{2x+2(2x+ 1)}*3^x)= (2^k)(3^2)\);

Equate the powers of 3 --> \(x=2\);
Equate the powers of 2 --> \(2x+2(2x+ 1)=k\) --> \(6x+2=k\) --> \(k=14\).

Answer: E.



Hi Bunuel

\((2^{2x+2(2x+ 1)}*3^x)= (2^k)(3^2)\); after this

\((2^{6x+2}*3^x)= (2^k)(3^2)\); \(x=2\); :? [color=#ff0000] where do you see power of 3 ? :? how you get 2 :?

how do you Equate the powers of 2 --> \(2x+2(2x+ 1)=k\) --> \(6x+2=k\) --> \(k=14\). and get 14 :?

i know this rule -->

if \(a^x = a^y\) then \(x = y\)

in other words if the bases match then exponents are equal.

please explain :) \(thank you ^ {1000}\)

:)


2^(2x+2(2x+ 1))*3^x= (2^k)*(3^2);

Equate the powers of 3 --> \(x=2\);

Next, equate the powers of 2: \(2x+2(2x+ 1)=k\) --> \(6x+2=k\). SINCE x = 2 (form above), then \(k=14\).
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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what  [#permalink]

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Re: If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what   [#permalink] 05 Feb 2019, 16:36
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