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Deconstructing the Question

We are given

\(0.375^3 = 2^a 3^b 6^c\)

and must find

\(b-a\)

First convert the decimal into a fraction.

\(0.375 = \frac{3}{8}\)

Step-by-step

Compute

\(0.375^3 = \left(\frac{3}{8}\right)^3\)

\(=\frac{3^3}{8^3}\)

Since

\(8 = 2^3\)

we get

\(8^3 = (2^3)^3 = 2^9\)

Thus

\(0.375^3 = \frac{3^3}{2^9}\)

\(= 3^3 \cdot 2^{-9}\)

Now rewrite the right side.

\(6 = 2 \cdot 3\)

\(6^c = 2^c 3^c\)

So

\(2^a 3^b 6^c = 2^a 3^b (2^c 3^c)\)

\(=2^{a+c} 3^{b+c}\)

Now match exponents with

\(2^{-9} 3^3\)

\(a+c = -9\)

\(b+c = 3\)

Subtract the equations

\((b+c)-(a+c) = 3 - (-9)\)

\(b-a = 12\)

Answer D: 12
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