已知 $a=\left(\dfrac 12\right)^b={\log_{0.3}}0.2$，则（ ）
A．$2.5<2a-b<3$
B．$3<2a-b<3.5$
C．$3.5<2a-b<4$
D．以上答案都不对
答案 B．
解析 首先列乘方表\[\begin{array}{c|cccccccc}\hline n&1&2&3&4&5&6&7&8\\ \hline 0.3^n&0.3&0.09&0.027&0.0081&0.00243&0.000729&0.0002187&0.00006561\\ \hline 0.2^n&0.2&0.04&0.008&0.0016&0.00032&0.000064&0.0000128&0.00000256 \\ \hline\end{array}\] 于是 $ \dfrac 43<{\log_{0.3}}0.2 <\dfrac 75$，再估计 $-b={\log_2}a$，再列乘方表\[\begin{array}{c|cccccc}\hline n&1&2&3&4&5&6\\ \hline 2^n&2&4&8&16&32&64\\ \hline \left(\dfrac 43\right)^n&\dfrac 43&\dfrac{16}9&\dfrac{64}{27}&\dfrac{256}{81}&\dfrac{1024}{243}&\dfrac{4096}{729}\\ \hline \left(\dfrac 75\right)^n&\dfrac75&\dfrac{49}{25}&\dfrac{343}{125}&\dfrac{3087}{625}&\dfrac{21606}{3125}&\\ \hline \end{array}\] 于是 $ \dfrac 25<{\log_2}\dfrac 43<{\log_2}a<{\log_2}\dfrac 75<\dfrac 12$． 综上所述，有 $ \dfrac83+\dfrac 25<2a-b<\dfrac {14}5+\dfrac 12$，也即 $\dfrac{46}{15}<2a-b<\dfrac{33}{10}$．