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R^c)(R^d)(R^e)=12^-12. If R>0, and c,d and e are each [#permalink]
 25 Jul 2003, 07:53
(R^c)(R^d)(R^e)=12^-12.
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) -12
B)-10
C)-9
D)-6
E)-1
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KL
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(R^c)(R^d)(R^e)=12^-12.
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) -12
B)-10
C)-9
D)-6
E)-1
My answer is (A) i.e. -12
(R^c)(R^d)(R^e)=12^-12
=> r^(c+d+e) = 12^-12
=> r^(c+d+e) = (2*sqrt3)^-24 (see below how I got this)
so, lowest value of c+d+e = -24
which leads to the smallest value of c as -12.
12^-12
= (4*3)^-12
= 4^-12 * 3^-12
= [(2^2)]^-12 * [(sqrt3) ^ 2] ^ -12
= 2^-24 * (sqrt3)^-24
= [2 * sqrt(3)] ^ -24
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If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
You are assuming that R is an interger >0. But in the problem c,d,e are defined as negative intergers. R may not be an integer - isn't it?
If we take, R=2√3; c=-12, d=-7, e=-5
then, c+d+e=-24
so, R^(c+d+e) = (2√3)^-24 = 12^-12
That's why I think answer is A. Feedback pls.
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prakuda2000 wrote: If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
You are assuming that R is an interger >0. But in the problem c,d,e are defined as negative intergers. R may not be an integer - isn't it?
If we take, R=2√3; c=-12, d=-7, e=-5
then, c+d+e=-24
so, R^(c+d+e) = (2√3)^-24 = 12^-12
That's why I think answer is A. Feedback pls.
But how can you 'guess'tify R to be a non-integer
Even if c+d+e = -24 , why C is -12 , why can't it be -22 ?
Per the question C, D and E need to be -negetive integers , So each one of them may atmost be -1.
So this gives us C to be -22 ?
I feel (A) should be - 22 (may be a typo ).
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Re: Some algebra practice [#permalink]
26 Jul 2003, 03:48
Lynov Konstantin wrote: (R^c)(R^d)(R^e)=12^-12.
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) -12
B)-10
C)-9
D)-6
E)-1
Sorry, guys. I made a typing mistake in the question.
The condition should be read as follows:
(R^c)(R^d)(R^e)=R^c+d+e= R^-12
If R>0, and c,d and e are each different negative integers, what is the smalest that c could be?
A) -12
B)-10
C)-9
D)-6
E)-1
Forgive me please and give it another shot.
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KL
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R^(a+b+c) = 12^(-12)
so a+b+c = -12
Given a,b and c are -ve distincive integers
abc can be any of(0,0,-12), (0,-1,-11),(-1,-1,-10),(-1,-2,-9)
Lowest is -11 but that choice is not given. Next lowest is
(-1,-2,-9) where integers are distinct and c = -9
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