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If 2a – 2−a = , what is the value of 4a + 4−a ?

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If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 08 Dec 2018, 07:37
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If \(2^a – 2^{−a}\)= \(\sqrt{7}\) , what is the value of\(4^a + 4^{−a}\) ?


A. 1
B. 7
C. 9
D. 49
E. It cannot be determined from the information given.
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Re: If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 08 Dec 2018, 07:56
1
Squaring \(2^a - 2^{-a}\) gives, \((2^a - 2^{-a})^2\) = \(4^a+4^{-a}-2*2^a*2^{-a}\)

\((\sqrt{7})^2\) = \(4^a+4^{-a}-2*2^a*2^{-a}\)

7 = \(4^a+4^{-a}-2*1\)

9 = \(4^a+4^{-a}\)

C is the answer.
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Re: If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 08 Dec 2018, 08:04
1
OA:C

\(2^a – 2^{−a}\)= \(\sqrt{7}\)

Squaring both sides, we get
\(2^{2a}+2^{−2a}-2*2^a*2^{-a}=7\)
\(4^a+4^{-a}-2=7\)
\(4^a+4^{-a}=9\)
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Re: If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 08 Dec 2018, 08:06
2
Scmfanatic wrote:
If \(2^a – 2^{−a}\)= \(\sqrt{7}\) , what is the value of\(4^a + 4^{−a}\) ?


A. 1
B. 7
C. 9
D. 49
E. It cannot be determined from the information given.


\(4^a + 4^{−a}\) = \(2^2a + 2^{−2a}\)

given,
\(2^a – 2^{−a}\)= \(\sqrt{7}\) ; squaring both sides
(a-b)^2 = a^2+b^2-2ab ; we get \(2^2a + 2^{−2a} - 2\) = 7
solving we get \(2^2a + 2^{−2a}\) = 9 C
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Re: If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 08 Dec 2018, 08:08
1
Important thing to note in this question is whenever you have a number and its reciprocal then remember that when ever you mutiply them would give you one. Also looking at the numbers you should note that (A + B)^2 = A^2 + B^2 + 2AB should do the trick in this question.

Option C is the correct answer.
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If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 05 Feb 2019, 20:27
Afc0892 wrote:
Squaring \(2^a - 2^{-a}\) gives, \((2^a - 2^{-a})^2\) = \(4^a+4^{-a}-2*2^a*2^{-a}\)

\((\sqrt{7})^2\) = \(4^a+4^{-a}-2*2^a*2^{-a}\)

7 = \(4^a+4^{-a}-2*1\)

9 = \(4^a+4^{-a}\)

C is the answer.


Good night Afc0892 !

Would you be so kind and explain to me how do you go from 2^a*2^{-a} to *1?

Kind regards!
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Re: If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 05 Feb 2019, 21:01
1
jfranciscocuencag wrote:
Afc0892 wrote:
Squaring \(2^a - 2^{-a}\) gives, \((2^a - 2^{-a})^2\) = \(4^a+4^{-a}-2*2^a*2^{-a}\)

\((\sqrt{7})^2\) = \(4^a+4^{-a}-2*2^a*2^{-a}\)

7 = \(4^a+4^{-a}-2*1\)

9 = \(4^a+4^{-a}\)

C is the answer.


Good night Afc0892 !

Would you be so kind and explain to me how do you go from 2^a*2^{-a} to *1?

Kind regards!


Hey jfranciscocuencag, sure :)

\(2^a*2^{-a}\) can be written as \(2^a*\frac{1}{2^a}\) then \(2^a\) will be cancelled out.

Hope it helps
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Re: If 2a – 2−a = , what is the value of 4a + 4−a ?  [#permalink]

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New post 07 Feb 2019, 19:44
Scmfanatic wrote:
If \(2^a – 2^{−a}\)= \(\sqrt{7}\) , what is the value of\(4^a + 4^{−a}\) ?


A. 1
B. 7
C. 9
D. 49
E. It cannot be determined from the information given.


We can square both sides of the given equation and obtain:

(2^a - 2^(-a))^2 = (√7)^2

4^a + 4^(-a) - 2 = 7

4^a + 4^(-a) = 9

Answer: C
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Re: If 2a – 2−a = , what is the value of 4a + 4−a ?   [#permalink] 07 Feb 2019, 19:44
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