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

600 level is okay.
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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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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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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.

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

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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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 --> $$2x+2(2x+ 1)=k$$ --> $$6x+2=k$$ --> $$k=14$$.

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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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

Given: $$(12^x)(4^{2x+ 1})= (2^k)(3^2)$$

Rewrite $$12$$ and $$4$$ as follows: $$[(2^2)(3^1)]^x[(2^2)^{2x+ 1}]= (2^k)(3^2)$$

Apply exponent laws to get: $$(2^{2x})(3^x)(2^{4x+ 2})= (2^k)(3^2)$$

On the left side, combine the powers of $$2$$ to get: $$(2^{6x+2})(3^x)= (2^k)(3^2)$$

At this point we can see that $$6x+2 = k$$ AND $$x = 2$$

Take: $$6x+2 = k$$

Replace $$x$$ with $$2$$ to get: $$6(2)+2 = k$$

Evaluate: $$14 = k$$

Cheers,
Brent
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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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$$12^x4^{(2x+1)}= (2^k)(3^2)$$
$$(3*2^2)^x(2^2)^{(2x+1)}= (2^k)(3^2)$$
$$3^x2^{x+2(2x+1)}= (2^k)(3^2)$$

$$x=2$$
$$k=2x+2(2x+1)$$
$$k=2*2+2(2*2+1)$$
$$k=14$$

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If x and k are integers and (12^x)(4^2x+1)= (2^k)(3^2), what [#permalink]
Asked: If x and k are integers and $$(12^x)(4^{2x+ 1})= (2^k)(3^2)$$, what is the value of k ?

$$3^x*2^{2x}*2^{4x+2} = 2^k*3^2$$
$$2^{6x+2}*3^x = 2^k*3^2$$

Equating the powers of 3 & 2 respectively, we get
x = 2
k = 6x+2 = 14

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