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

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Founder
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If 2^2p-10=2(2^9-2p), then p= [#permalink]  21 Nov 2018, 08:24
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Question Stats:

92% (00:57) correct 7% (03:34) wrong based on 13 sessions
If $$2^{2p-10}=2(2^{9-2p})$$, then p=

A. 5

B. 3

C. 0

D. -1

E. -2
[Reveal] Spoiler: OA

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Re: If 2^2p-10=2(2^9-2p), then p= [#permalink]  23 Nov 2018, 12:33
1
KUDOS
Carcass wrote:
If $$2^{2p-10}=2(2^{9-2p})$$, then p=

A. 5

B. 3

C. 0

D. -1

E. -2

Easiest way is to back-substitute.

p = 5

2^(10-10) = 2 (2^(9 - 10))

2^0 = 2* 2^-1

2^0 = 2/2

1 = 1

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Re: If 2^2p-10=2(2^9-2p), then p= [#permalink]  27 Nov 2018, 06:53
it was the same question as in the exam
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Re: If 2^2p-10=2(2^9-2p), then p= [#permalink]  27 Nov 2018, 06:54
can you provide the exmpalantion
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Re: If 2^2p-10=2(2^9-2p), then p= [#permalink]  10 Feb 2019, 09:20
This question tests the rules of exponents.

Remember that a^x * a^y = a^(x+y)

Sometimes they also create a question with a^x * a^x = (?) to try to trick you. Just follow the rule above. [a^x * a^x = a^(x+x)]
Re: If 2^2p-10=2(2^9-2p), then p=   [#permalink] 10 Feb 2019, 09:20
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