College Algebra, Exercise P, Exercise P.4, Section Exercise P.4, Problem 26

Solve the expression $\displaystyle \frac{3^2 \cdot 4^{-2} \cdot 5}{2^{-4} \cdot 3^3 \cdot 25 }$

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\begin{equation}
\begin{aligned}
\frac{3^2 \cdot 4^{-2} \cdot 5}{2^{-4} \cdot 3^3 \cdot 25 } &= \frac{2^4\cdot 3^2\cdot 5}{4^2\cdot 3^3\cdot 25} && \text{Law: } \frac{a^{-n}}{b^{-m}} = \frac{b^m}{a^n}\\
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&= \frac{2^4\cdot 3^2 \cdot 5}{(2^2)^2 \cdot 3^3 \cdot5^2} && \text{Factor}\\
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&= \frac{2^4 \cdot 3^2 \cdot 5}{2^4 \cdot 3^3 \cdot 5^2} && \text{Law: } \frac{a^m}{a^n} = a^{m-n}\\
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&= 2^{4-4} \cdot 3^{2-3} \cdot 5^{1-2} && \text{Simplify}\\
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&= 2^0 \cdot 3^{-1} \cdot 5^{-1} && \text{Definition of negative exponents } a^{-n} = \frac{1}{a^n}\\
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&= \frac{1}{3 \cdot 5} && \text{Simplify}\\
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&= \frac{1}{15}
\end{aligned}
\end{equation}
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