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2^x - 2^(x-2) = 3(12^13) a) 9 b) 11 c) 13 d) 15 e) 17

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SVP
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2^x - 2^(x-2) = 3(12^13) a) 9 b) 11 c) 13 d) 15 e) 17  [#permalink]

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New post 07 Nov 2008, 17:16
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2^x - 2^(x-2) = 3(12^13)

a) 9
b) 11
c) 13
d) 15
e) 17





I had this written in my notes ages ago. However, I'm starting to doubt whether I copied this question correctly. If someone can answer this correctly, then I would be certain that my notes is correct, and would need an explanation.
Thanks

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Re: PS: exponents [#permalink]

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New post 07 Nov 2008, 17:44
2^x - 2^(x-2) = 3(12^13)

<=> 2^x ( 1 - 1/4) = 3*(12^13)
<=> 2^x* 3/4 = 3*12^13
<=> 2^x = 4*12^13
<=> 2^x = 2^2 * 2^(2*13) * 3^13
<=> 2^x = 2^18 * 3^13

So your note must be incorrect, if 2^x - 2^(x-2) = 3(2^13), the answer will be x = 15 (D)

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Re: PS: exponents [#permalink]

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New post 07 Nov 2008, 18:47
I could not solve the way you wrote it down.

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Re: PS: exponents [#permalink]

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New post 08 Nov 2008, 02:57
lylya4 wrote:
2^x - 2^(x-2) = 3(12^13)

<=> 2^x ( 1 - 1/4) = 3*(12^13)
<=> 2^x* 3/4 = 3*12^13
<=> 2^x = 4*12^13
<=> 2^x = 2^2 * 2^(2*13) * 3^13
<=> 2^x = 2^18 * 3^13

So your note must be incorrect, if 2^x - 2^(x-2) = 3(2^13), the answer will be x = 15 (D)



yeah, exactly. The answer I have in my notes is D=15. So instead of writing 2^13, i must have accidentally wrote 12^13.

thanks

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Re: PS: exponents   [#permalink] 08 Nov 2008, 02:57
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2^x - 2^(x-2) = 3(12^13) a) 9 b) 11 c) 13 d) 15 e) 17

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