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From the question, we can deduce the answer using indices which deals with power of numbers, e.g
[tex]\frac{5^2.5^3}{5^4}=\frac{5^{2+3}}{5^4}=\frac{5^5}{5^4}=5^{5-4}=5^1=5[/tex]Also, for the next one,
[tex]\frac{9^4.9^6}{9^8}=\frac{9^{4+6}}{9^8}=\frac{9^{10}}{9^8}=9^{10-8}=9^2[/tex]This applies to every other ones provided in the question,
Considering Option D,
[tex]\begin{gathered} \frac{b^{2k}.b^{3k}}{b^{4k}} \\ \text{where k is 2,} \\ \frac{b^{2(2)}.b^{3(2)}}{b^{4(2)}}=\text{ }\frac{b^4.b^6}{b^8}=\frac{b^{4+6}}{b^8}=\frac{b^{10}}{b^8}=b^{10-8}=b^2 \\ \text{Where k=2, Option D is favoured} \end{gathered}[/tex]Since the equation D favours the solution,
Hence, the best option is D.