当n趋向于无限大时，n的0.1次方和logn谁大

起因是和一个帅哥（sb）讨论另一个问题，延伸到这个问题上面来的

怀着 对问题的严谨性我打开了尘封已久的pycharm

代码：

import math

if __name__ == '__main__':
    x = 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000;
    print(math.log(x,2));
    print(x ** 0.01);

控制台输出：

 发现只要数足够大，那么n的0.001次方就大于logn