If $a, b, c$ are non-zero and $14^a = 36^b = 84^c$, then $6b \left( \frac{1}{c} - \frac{1}{a} \right)$ is equal to

We need to find the value of $6b \left( \tfrac{1}{c} - \tfrac{1}{a} \right)$ given that $14^a = 36^b = 84^c$.

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Since all three expressions are equal, let's call this common value $K$:

$14^a = 36^b = 84^c = K$

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From the equation above, we can write:

$14 = K^{1/a}$ (taking the $a$-th root of both sides)

$36 = K^{1/b}$ (taking the $b$-th root of both sides)

$84 = K^{1/c}$ (taking the $c$-th root of both sides)

Why does this work? If $14^a = K$, then $14 = K^{1/a}$ because $(K^{1/a})^a = K$.

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Here's the key insight: Let's see how these numbers relate to each other.

Notice that $84 ÷ 14 = 6$ and $36 = 6^2$

This means: $(84 ÷ 14)^2 = 6^2 = 36$

Or written differently: $\left(\tfrac{84}{14}\right)^2 = 36$

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Replacing with our $K$ expressions:

$\left(\tfrac{K^{1/c}}{K^{1/a}}\right)^2 = K^{1/b}$

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Using the rule $\tfrac{x^m}{x^n} = x^{m-n}$:

$\left(K^{1/c - 1/a}\right)^2 = K^{1/b}$

Using the rule $(x^m)^n = x^{mn}$:

$K^{2(1/c - 1/a)} = K^{1/b}$

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Since the bases are the same, the exponents must be equal:

$2\left(\tfrac{1}{c} - \tfrac{1}{a}\right) = \tfrac{1}{b}$

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$2b\left(\tfrac{1}{c} - \tfrac{1}{a}\right) = 1$

$6b\left(\tfrac{1}{c} - \tfrac{1}{a}\right) = 3$

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Therefore, the answer is 3.

Key takeaway: When dealing with equations like $a^x = b^y = c^z$, setting them equal to a common value $K$ and expressing each base in terms of $K$ often reveals useful relationships between the variables.