If $3^{a}=4,4^{b}=5,5^{c}=6,6^{d}=7,7^{e}=8$ and $8^{f}=9$, then the value of the product $\textit{abcdef }$ is

We need to find the product $abcdef$ when we have a chain of exponential equations. The key insight is that each equation connects to the next - the result of one equation becomes the base of the next equation!

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We start with: $8^f = 9$

Since $8 = 7^e$, we can substitute this into our equation:

$(7^e)^f = 9$

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When we have $(a^m)^n$, this equals $a^{mn}$

So: $7^{ef} = 9$

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Since $6^d = 7$, we substitute:

$6^{def} = 9$

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Since $5^c = 6$, we substitute:

$5^{cdef} = 9$

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Since $4^b = 5$, we substitute:

$4^{bcdef} = 9$

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Since $3^a = 4$, we substitute:

$3^{abcdef} = 9$

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Now we need to express 9 as a power of 3:

$9 = 3 \times 3 = 3^2$

So our equation becomes:

$3^{abcdef} = 3^2$

When we have the same base on both sides of an equation, the exponents must be equal.

$\therefore abcdef = 2$