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If x and k are integers and (12^x) (4^2^x^+^1) then K is ?
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04 Sep 2018, 22:25
04 Sep 2018, 22:25
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Question Stats:
88% (01:14) correct
11% (00:02) wrong based on 9 sessions
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If x and k are integers and \((12^x)(4^{2x+1})= 2^k \times 3^2\) then K is ?
A)4
B)5
C)10
D)14
E)15
src; orbit test prep
A)4
B)5
C)10
D)14
E)15
src; orbit test prep
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: If x and k are integers and (12^x) (4^2^x^+^1) then K is ?
[#permalink]
10 Apr 2020, 14:42
10 Apr 2020, 14:42
1
amorphous wrote:
If x and k are integers and \((12^x)(4^{2x+1})= 2^k \times 3^2\) then K is ?
A)4
B)5
C)10
D)14
E)15
src; orbit test prep
A)4
B)5
C)10
D)14
E)15
src; orbit test prep
Given: \((12^x)(4^{2x+1})= 2^k \times 3^2\)
Rewrite \(12\) and \(4\) as follows: \((2^23^1)^x(2^2)^{2x+1})= 2^k \times 3^2\)
Apply the Power of a Product Law to get: \((2^{2x}3^{x})(2^{4x+2})= 2^k \times 3^2\)
Simplify to get: \((2^{6x+2})(3^{x})= 2^k \times 3^2\)
When we compare powers of 3, we can see that: \(x = 2\)
Now take: \(2^{6x+2}= 2^k\)
Replace \(x\) with \(2\) to get: \(2^{6(2)+2}= 2^k\)
Simplify: \(2^{14}= 2^k\)
So, \(k=14\)
Answer: D
Cheers,
Brent
gmatclubot
Re: If x and k are integers and (12^x) (4^2^x^+^1) then K is ? [#permalink]
10 Apr 2020, 14:42
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