If a^{2x} = b^{3y} = c^{5z} and b = a^{2}c, then which one of the following is true?

Correct — C, 1/x + 1/5z = 1/3y. Give the three equal powers a common name: let a^(2x) = b^(3y) = c^(5z) = k. Then a = k^(1/2x), b = k^(1/3y) and c = k^(1/5z). Substitute these into the second relation, b = a²c. The right-hand side is (k^(1/2x))² x k^(1/5z) = k^(1/x) x k^(1/5z) = k^(1/x + 1/5z), and the left-hand side is k^(1/3y). Two equal powers of the same base have equal exponents, so 1/x + 1/5z = 1/3y, which is exactly option (c). The step that does the work is squaring a: the exponent 1/2x doubles to 1/x, and the 2 disappears — which is why no 2 survives in the correct relation.

• (a)2/x + 3/y = 5/z — Reads the exponents 2x, 3y and 5z straight off the first equation and inverts them, ignoring the relation b = a²c altogether. Without that second condition there is nothing to link the three quantities.
• (b)1/x − 3/5z = 1/3y — Has the correct 1/3y on the right but the wrong middle term. Substituting b = a²c produces a sum of 1/x and 1/5z, not a difference, and the coefficient on the last term is 1, not 3.
• (d)2/x − 5/z = 3/y — Keeps the numerators 2, 5 and 3 from the original exponents and joins them with a subtraction. Taking logarithms shows the coefficients change: squaring a turns 1/2x into 1/x, so the 2 cancels rather than surviving.

When several expressions are set equal to one another, naming the common value is almost always the first move. Writing a^(2x) = b^(3y) = c^(5z) = k converts three variables into three powers of one base, and any relation among a, b and c then becomes a linear relation among the reciprocal exponents. Taking logarithms achieves the same thing: 2x ln a = 3y ln b = 5z ln c, and b = a²c gives ln b = 2 ln a + ln c.

Every wrong option here is built from the numbers 2, 3 and 5 in the original exponents, which is precisely what the substitution changes. Watch the coefficient carefully: a raised to the power 1/2x, then squared, gives 1/x — the 2 in the exponent of a² cancels against the 2 in 2x. Once you see that no 2 can survive on the a term, options (a) and (d) are gone on sight, and the choice is between a sum and a difference. Since b = a²c is a product, the exponents add.

• Setting a^(2x) = b^(3y) = c^(5z) = k gives a = k^(1/2x), b = k^(1/3y), c = k^(1/5z).
• A product of powers of the same base adds exponents, so a²c becomes k^(1/x + 1/5z).
• Equating exponents on b = a²c gives 1/3y = 1/x + 1/5z.
• The 2 in a² cancels the 2 in 2x, so no factor of 2 survives in the final relation.
• Taking logarithms gives the same result: ln b = 2 ln a + ln c with ln a = ln k /(2x) and so on.

• Inverting the exponents of the first equation and ignoring the second relation.
• Carrying the factor of 2 into the answer when squaring has already cancelled it.
• Turning the product a²c into a difference of exponents rather than a sum.

An indices item where every wrong option is assembled from the numbers printed in the question.

No directly related past PYQ was found.

• practice — not a real PYQ

  If 2^x = 3^y = 6^z, then which relation holds?

  • (a)1/x + 1/y = 1/z
  • (b)1/x − 1/y = 1/z
  • (c)x + y = z
  • (d)xy = z

  Answer(a) 1/x + 1/y = 1/z — because 6 = 2 x 3, the reciprocal exponents add.

• practice — not a real PYQ

  If a^p = b^q and b = a³, then

  • (a)p = 3q
  • (b)q = 3p
  • (c)p = q
  • (d)pq = 3

  Answer(a) p = 3q — substituting b = a³ gives a^p = a^(3q), so the exponents are equal.