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# If 2^(2p-10) = 2(2^9-2p), then p =

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If 2^(2p-10) = 2(2^9-2p), then p = [#permalink]

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04 Oct 2013, 00:01
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If 2^(2p-10) = 2*2^(9-2p), then p =

A. 5
B. 3
C. 0
D. -1
E. -2
[Reveal] Spoiler: OA

Last edited by Bunuel on 04 Oct 2013, 00:15, edited 2 times in total.
Edited the question.

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Re: If 2^2p-10 = 2(2^9-2p), then p = [#permalink]

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04 Oct 2013, 00:06
avohden wrote:
If $$2 ^2^p^-^1^0 = 2(2 ^9^-^2^p)$$, then p =

A. 5
B. 3
C. 0
D. -1
E. -2

$$2 ^2^p^-^1^0 = 2(2 ^9^-^2^p)$$
this can be written as
$$2 ^2^p^-^1^0 = (2 ^10^-^2^p)$$
now equating the powers
$$2p-10=10-2p$$
$$4p = 20$$
$$p = 5$$

hence A
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Re: If 2^2p-10 = 2(2^9-2p), then p = [#permalink]

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04 Oct 2013, 00:09
avohden wrote:
If $$2 ^2^p^-^1^0 = 2(2 ^9^-^2^p)$$, then p =

A. 5
B. 3
C. 0
D. -1
E. -2

$$2 ^2^p^-^1^0 = 2(2 ^9^-^2^p)$$

2(2^(9-2p)) can be written as (2^1)*(2^(9-2p))=>2^(1+9-2p)=>2^(10-2p)

hence 2^(2p-10)=2^(10-2p)
2p-10=10-2p
4p=20
p=5
Ans 'A'

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Re: If 2^2p-10 = 2(2^9-2p), then p = [#permalink]

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04 Oct 2013, 00:14
avohden wrote:
If $$2 ^2^p^-^1^0 = 2(2 ^9^-^2^p)$$, then p =

A. 5
B. 3
C. 0
D. -1
E. -2

$$2 ^2^p^-^1^0 = 2^1^+^9^-^2^p$$ (adding the powers with same base)
$$2 ^2^p^-^1^0 = 2^1^0^-^2^p$$

Comparing the powers, (Since same base), we get

2p - 10 = 10 - 2p

p =5

Hope it helped.
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Re: If 2^(2p-10) = 2(2^9-2p), then p = [#permalink]

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26 Feb 2014, 03:21
Divide the equation by 2 & equate the powers

2p - 11 = 9 - 2p

4p = 20

p = 5 = Answer = A
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Re: If 2^(2p-10) = 2(2^9-2p), then p = [#permalink]

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04 May 2015, 22:07
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Re: If 2^(2p-10) = 2(2^9-2p), then p =   [#permalink] 04 May 2015, 22:07
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