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CA Foundation · Quantitative Aptitude · Ratio and Proportion, Indices and Logarithms

If a^x = b, b^y = c and c^z = a, where a, b, c are positive and different from 1, then the value of xyz is:

The value of xyz is 1. Substituting c = b^y into a = c^z gives a = b^(yz), and substituting b = a^x gives a = a^(xyz). As a is positive and not equal to 1, the exponents must match, so xyz = 1.

  1. A1Correct
  2. B0
  3. Cabc
  4. Dx + y + z

Explanation

Substitute successively: a = c^z = (b^y)^z = b^(yz) = (a^x)^(yz) = a^(xyz). Since a is positive and not 1, the exponents must be equal, so xyz = 1. Zero would make a = 1, which is excluded.

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