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by everything's eventual » Tue Sep 04, 2012 7:11 pm
a^x = b
b^y = c
c^z = a

Multiply both sides and we get the following :

(a^x)* (b^y)* (c^z) = abc

This equation can be true only if x,y and z are all 1.

Therefore xyz = 1.
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by everything's eventual » Tue Sep 04, 2012 7:13 pm
a^x = b
b^y = c
c^z = a

Multiply both sides and we get the following :

(a^x)* (b^y)* (c^z) = abc

This equation can be true only if x,y and z are all 1.

Therefore xyz = 1.
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by GMATGuruNY » Wed Sep 05, 2012 12:47 am
shanice wrote:If a, b, and c are not equal to 0 or 1 and if a^x = b, b^y = c, and c^z = a, then xyz =
(A) 0
(B) 1
(C) 2
(D) a
(E) abc

This question is pretty challenging for me.please help me out here.
PLUG AND CHUG.

Let c=2 and z=1.
Then a = c^z = 2¹ = 2.

Since b^y = c and c=2, we get:
b^y = 2.
Let b=2 and y=1.

Since a^x = b, a=2, and b=2, we get:
2^x=2.
Thus, x=1.

xyz = 1*1*1 = 1.

The correct answer is B.
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by vk_vinayak » Wed Sep 05, 2012 3:07 am
shanice wrote:If a, b, and c are not equal to 0 or 1 and if a^x = b, b^y = c, and c^z = a, then xyz =
(A) 0
(B) 1
(C) 2
(D) a
(E) abc

This question is pretty challenging for me.please help me out here.
a^x = b, b^y = c, and c^z = a

Start with a^x = b. ------- (1)

But it is given that a = c^z. Substitue the value of a in (1). We get (c^z)^x = b => (c)^zx = b --- (2)

We are also given that c = b^y. Substitute this value of c in (2).

(c)^zx = b
(b^y)^zx = b
(b)^yzx = b

(b)^yzx = (b)^1. So, we get xyz=1.
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by hjafferi » Wed Sep 05, 2012 5:54 am
B

If a^x = b then a = b^(1/x) similarly b= c^(1/y) and c=a^(1/z)

Put value of c in equation 2
b=[a^(1/z)]^(1/y)
Therefore b= a^(1/(ya))

But b= a^x then
A^x=a^(1/(ya))
Taking root x on both sides
A=a^(1/(xyz))

Since bases are same powers would have to be same
1/(xyz) = 1
Xyz=1

Hence B is the correct answer
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