楼主
5^0.6与log_2(5）怎样比较大小？
有没有简单易行的方法？1楼
首先,2<log2(5),5^0.6<3;
$$5^0.6=5^(3/5),
2^5=32<5^3=125<3^5=273;
2^2=4<5<2^3=8$$
其次,log2(5)<2.5<5^0.6;
$$5<4*1.414=2^2.5;
2.5^5=(5^5)/(2^5)=3175/32<125=5^3$$
所以,log2(5)<5^0.6
$$5^0.6=5^(3/5),
2^5=32<5^3=125<3^5=273;
2^2=4<5<2^3=8$$
其次,log2(5)<2.5<5^0.6;
$$5<4*1.414=2^2.5;
2.5^5=(5^5)/(2^5)=3175/32<125=5^3$$
所以,log2(5)<5^0.6
2楼
写得太乱啦，说白了，log2(5)<2.5<5^0.6
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