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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a

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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a  [#permalink]

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New post 14 Sep 2011, 10:31
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Does 4^a = 4^-a + b?

(1) 16^a = 1 + ((2^2a) / (b^-1))
(2) a = 2

Spoiler: :: Doubt
Hello, I'm having trouble understanding why statement 1 is sufficient.

In the explanation, statement 1 is simplified to 4^2a = 1 + (4^a)b and then both sides are divided by 4^a.

Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner?

Thank you.
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Re: MGMAT Advanced quant #4-16 (pg 118)  [#permalink]

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New post 14 Sep 2011, 10:59
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rjdunn03 wrote:

Wouldn't 4^2a divided by 4^a be 4^2?


From the exponent rules, if you divide two numbers with exponents, you subtract the powers. So, for example, 2^9/2^4 = 2^(9-4) = 2^5. In the same way, 4^(2a)/4^a = 4^(2a - a) = 4^a; it is not equal to 4^2 (unless a is equal to 2).

I'm just running out the door, so I don't even have time to actually look at the question, but I can later if you still aren't clear on the solution.
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Re: MGMAT Advanced quant #4-16 (pg 118)  [#permalink]

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New post 14 Sep 2011, 11:19
IanStewart wrote:
rjdunn03 wrote:

Wouldn't 4^2a divided by 4^a be 4^2?


From the exponent rules, if you divide two numbers with exponents, you subtract the powers. So, for example, 2^9/2^4 = 2^(9-4) = 2^5. In the same way, 4^(2a)/4^a = 4^(2a - a) = 4^a; it is not equal to 4^2 (unless a is equal to 2).

I'm just running out the door, so I don't even have time to actually look at the question, but I can later if you still aren't clear on the solution.



Oh yes, thank you. I understand now, I wasn't thinking it through properly.

I appreciate your help.
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Re: MGMAT Advanced quant #4-16 (pg 118)  [#permalink]

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New post 14 Sep 2011, 12:59
Statement 1 is only sufficient ....We can solve for a and b values....

Statement 2 in suff..

IMO A
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Re: MGMAT Advanced quant #4-16 (pg 118)  [#permalink]

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New post 15 Sep 2011, 00:27
rjdunn03 wrote:
Does 4^a = 4^-a + b?
Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner?
Thank you.


no it wouldnt. 4^2a/4^a= 4^(2a-a)=4^a

rjdunn03 wrote:
Does 4^a = 4^-a + b?

(1) 16^a = 1 + ((2^2a) / (b^-1))
(2) a = 2

Thank you.

we have 4^a = 4^-a + b or 4^2a = 1+4^a*b


stmt 1-
16^a = 1 + ((2^2a) / (b^-1))
4^2a=1+4^a*b
suff

stmt 2 is insuf, cuz b could be anything
OA is A
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Re: MGMAT Advanced quant #4-16 (pg 118)  [#permalink]

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New post 15 Sep 2011, 16:59
2
4^a = 4^(-a) + b?

rephrasing we have 4^a = (1/(4^a)) + b?

=> (4^a)(4^a) = 1 + (4^a)*b?

=> 16^a = 1 + (4^a)*b?


1. Sufficient

16^a = 1+ (4^a)/(b^-1)

= 1+(4^a)*b

2. Not sufficient.

we dont know anything about b.

Answer is A.
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Re: Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a  [#permalink]

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New post 26 Nov 2014, 15:00
rjkaufman21 wrote:
Does 4^a = 4^-a + b?

(1) 16^a = 1 + ((2^2a) / (b^-1))
(2) a = 2

Spoiler: :: Doubt
Hello, I'm having trouble understanding why statement 1 is sufficient.

In the explanation, statement 1 is simplified to 4^2a = 1 + (4^a)b and then both sides are divided by 4^a.

Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner?

Thank you.



I did it in the following way:

let \(4^a = t\), then the question becomes, is \(t^2-tb-1 = 0\)?

from statement 1:\(t^2 = 1 + tb\), or\(t^2-tb-1 = 0\), the required info
from statement 2: we cannot tell if \(t^2-tb-1 = 0\), so NSF

Answer [A]
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Re: Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a  [#permalink]

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New post 12 May 2015, 03:42
Is it correct to simplify (4^a)*(4^a) as (4^a)^2 and then as 4^2a which equals 16^a?
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Re: Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a  [#permalink]

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New post 12 May 2015, 03:52
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Re: Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a  [#permalink]

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New post 22 Nov 2016, 09:02
rjkaufman21 wrote:
Does 4^a = 4^-a + b?

(1) 16^a = 1 + ((2^2a) / (b^-1))
(2) a = 2

Spoiler: :: Doubt
Hello, I'm having trouble understanding why statement 1 is sufficient.

In the explanation, statement 1 is simplified to 4^2a = 1 + (4^a)b and then both sides are divided by 4^a.

Wouldn't 4^2a divided by 4^a be 4^2? Or what am I missing? How does it simplify in such a manner?

Thank you.



B is clearly insufficient.

we can rewrite 4^a = 4^-a + b as: 2^2a = 2^-2a +b
or
(2^4a - 1)/2^2a = b

1. says exactly this.
(2^2a) / (b^-1) = 2^2a * b
16^a = 2^4a
2^4a - 1 = 2^2a*b
b= (2^4a - 1)/2^2a

sufficient.
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Re: Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a  [#permalink]

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Re: Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a   [#permalink] 19 Mar 2019, 00:26
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