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At the start of a project, a research lab keeps records in two separate digital archives. Archive M begins with 2 records and quadruples in size every 4 days. Archive N begins with 16 records and doubles in size every 6 days. If each archive grows continuously according to its stated pattern, how many days after the start will the two archives hold the same number of records?
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Correct answer
B
Translate each archive into a power of the same base, then set the two expressions equal.
Method (common base): Write both starting amounts and both growth factors as powers of 2. Archive M starts at 2, which is 2¹, and quadruples every 4 days. Because quadrupling is the same as two doublings, growing by a factor of 4 every 4 days means multiplying by 2 twice every 4 days, so after t days Archive M has 2⁽¹ ⁺ ᵗ/²⁾ records. Archive N starts at 16, which is 2⁴, and doubles every 6 days, so after t days it has 2⁽⁴ ⁺ ᵗ/⁶⁾ records.
Setting the two equal and matching exponents on the common base: 1 + t/2 = 4 + t/6. Multiply through by 6: 6 + 3t = 24 + t, so 2t = 18 and t = 9.
Check: at t = 9, Archive M has 2⁽¹ ⁺ ⁴·⁵⁾ = 2⁵·⁵, and Archive N has 2⁽⁴ ⁺ ¹·⁵⁾ = 2⁵·⁵. They match, so the answer is (B).
Method (doubling-gap reasoning): Archive N starts 8 times as large as Archive M, and 8 = 2³, a gap of three doublings. Archive M doubles every 2 days while Archive N doubles every 6 days, so Archive M gains ground at a rate of ½ − ⅙ = ⅓ of a doubling per day. To close three doublings takes 3 ÷ ⅓ = 9 days, confirming (B).
Why each wrong choice is tempting: (A) 6 comes from treating Archive N as fixed and dropping its growth term (solving 1 + t/2 = 4). (C) 12 comes from losing Archive M's starting amount of 2 in the setup (solving t/2 = 4 + t/6). (D) 18 comes from misreading Archive N's doubling interval as 3 days instead of 6 (solving 1 + t/2 = 4 + t/3). (E) 36 comes from rewriting the quadrupling in base 2 but using the exponent t/4 instead of t/2, forgetting that one quadrupling equals two doublings (solving 1 + t/4 = 4 + t/6). Each of these is a single specific slip in either the word-to-exponent translation or the same-base conversion, while (B) applies both correctly.
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