Let $3^{a} = 4, 4^{b} = 5, 5^{c} = 6, 6^{d} = 7, 7^{e} = 8$ and $8^{f} = 9$ . What is the value of the product abcdef?

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Solution

The correct option is B
2

$8^{f} = 9 \to 8 = 9^{\frac{1}{f}}$
$7^{e} = 8 \to 7^{e} = 9^{\frac{1}{f}} \to 7 = 9^{\frac{1}{e f}}$
Continuing in the same way we get $3 = 9^{\frac{1}{a b c d e f}}$
$⟹ 3 = 3^{\frac{2}{a b c d e f}}$
$⟹ \frac{2}{a b c d e f} = 1 \Rightarrow a b c d e f = 2$

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