logx < x < x*logx < x^2 < 2^x
- Each part of the above statement is true beyond a certain value of x and
- All of the above statement is also true for x > 2 (if log is taken at base 2)
- Now what’s missing above is how does 2 ^ logx compare to the above values? Hmmm how about an image?
- Here’s how, below is the diagram comparing 2 ^ logx to the rest.(Octave code). Remember a ^ log b is same as b ^ log a hence 2 ^ logx is same as x ^ log2. which is less than x since log2 to the base e < 1.
logx < 2 ^ logx < x < x*logx < x^2 < 2^x
