2024/25 AHSMC Part 1 Problem 9

Problem 9 of 16

$a$ , $b$ and $c$ are positive real numbers with the property that $2^a = 3^b = 216^c$ . Find the value of $\dfrac{ab}{bc+ca}$ .

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Solution

We have $6^{ab} = 2^{ab}3^{ab} = (2^a)^b(3^b)^a = (216^c)^{b+a} = 6^{3(cb+ca)},$ so $ab = 3(bc+ca)$ and $\dfrac{ab}{bc+ca} = 3$ . The answer is (E).

Alternatively, $c = a\log_{216} 2 = b\log_{216} 3$ , and $\frac{ab}{bc+ca} = \frac{1}{\log_{216} 2 + \log_{216} 3} = \frac{1}{\log_{216} 6} = \log_6 216 = \log_6(6^3) = 3.$

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