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655-705 (Hard)|   Algebra|      
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rjkaufman21
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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 14 Sep 2011, 10:31
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A
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55% (hard)

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62% (02:11) correct 38% (02:36) wrong based on 290 sessions

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Does 4^a = 4^-a + b?

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

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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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A
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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 14 Sep 2011, 10:59
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rjdunn03

Wouldn't 4^2a divided by 4^a be 4^2?
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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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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 14 Sep 2011, 11:19
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rjdunn03

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

I appreciate your help.
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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 14 Sep 2011, 12:59
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Statement 1 is only sufficient ....We can solve for a and b values....

Statement 2 in suff..

IMO A
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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 15 Sep 2011, 00:27
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rjdunn03
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.
Show more

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

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

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

Thank you.
Show more
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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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 15 Sep 2011, 16:59
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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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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 26 Nov 2014, 15:00
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rjkaufman21
Does 4^a = 4^-a + b?

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

Show SpoilerDoubt
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.
Show more


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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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 12 May 2015, 03:42
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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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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 12 May 2015, 03:52
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noTh1ng
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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Yes.

You could also do (4^a)*(4^a) = 4^(a+a) = 4^(2a) = 16^a.
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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 22 Nov 2016, 09:02
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rjkaufman21
Does 4^a = 4^-a + b?

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

Show SpoilerDoubt
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.
Show more


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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Does 4^a = 4^-a + b? (1) 16^a = 1 + ((2^2a) / (b^-1)) (2) a 19 Jul 2025, 18:13
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