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# If a, b, and c, are non-zero integers, is a^b = a^c ? (1)

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If a, b, and c, are non-zero integers, is a^b = a^c ? (1) [#permalink]

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27 Sep 2005, 04:17
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If a, b, and c, are non-zero integers, is a^b = a^c ?

(1) a^3 = a^4

(2) b^a = c^a
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27 Sep 2005, 06:06
I choose A.

1. suff. a^3=a^4 means that a must equal 1, therefore 1^b=1^c no matter what b and c are. Am I correct in this thinking?

2. insuff. if a was odd then b=c, but if a is even then b=c, but it b=-c, or -b=c. for instance a=3, b=-2, then c=-2, b^a=c^a would give you the answer to the stem. But if a=2, then b=-2 and c=2 or -2, which would not answer the stem.

What is OA?
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27 Sep 2005, 06:11
A for me

with statement 1 we know that A can only be 0 or 1
then we know that a^b = a^c = 1 or 0

with statement 2, it can be (-2)^2=2^2=4
however in that case a^-2 will differ from a^2
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27 Sep 2005, 07:18
OA is A...ciao Ant..good job=)
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27 Sep 2005, 07:36
1) a = 1 so suff.
2) insuff.

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28 Sep 2005, 03:33
Macedon wrote:
If a, b, and c, are non-zero integers, is a^b = a^c ?

(1) a^3 = a^4

(2) b^a = c^a

From statement 1, a = 1 so a ^ b = a ^ c. But statement 2 is insufficient. Hence the answer should be A.
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28 Sep 2005, 04:59
Antmavel wrote:
A for me

with statement 1 we know that A can only be 0 or 1
then we know that a^b = a^c = 1 or 0
...

We are also given a is non-zero so a=1 is only solution.

I always try and rearrange into standard equation form.

a^4 = a^3
a^4 - a^3 = 0
(a-1)(a-0)(a-0)(a-0) = 0
a=1, a=0 are roots
We are given a<>0 so a=1

For (2)
b^a = c^a
So (b/c)^a = 1 (as c<>0)
Now a<>0, so
(b/c) is a real root of 1.
So b/c = +/- 1
So b = +/-c

a^b = a^c if b=+c
a^b <> a^c if b=-c
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28 Sep 2005, 07:28
richardj wrote:

We are also given a is non-zero so a=1 is only solution.

right. thanks
28 Sep 2005, 07:28
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