how can i prove that x^(1/2)>ln(x) ?

Question

root(x) > natural log (x)?

Answer 1

The natural log is only defined when x>0 -- so ignore zero and the negative numbers.

This is obviously true when x < 1 -- since LN(x) is negative and x^(1/2) is positive.

We can rewrite this as:

prove that x^(1/2) - ln(x) > 0

We know that it is positive for x <= 0. If we could prove that the function x^(1/2) - ln(x) is strictly increasing, that would prove that the inequality holds.

To do this, we need only show that the first derivative is positive.

The first derivative is:

0.5*x(-0.5) - 1/x

so we need to show that

0.5*x(-0.5) - 1/x > 0

This is the same as

0.5*x(-0.5) > 1/x

Since X is positive, this is the same as

0.5*x^0.5 > 1

For x>1, this is always true -- so the function is always increasing -- which means the original inequality holds.

Answer 2

very carefully.
+2 pts. nooch