You can’t just remove the ½, and your method would only show for some value of N, for x>N would ln(x) < x^1/2Vardd said:Hi Grossgermany,
To answer the question, I'm assuming you have a rough idea of Calculus.
We know that the derivative of a formula, tells us the rate of change of the formula. We also that both formulas are always positive.
(Sorry no Latex I tried to make it works, but i failed. )
So let's take the derivative of ln(x) d/dx (ln(x)) = 1/x
Then of sqrt(x) d/dx (sqrt(x)) = 1/(2sqrt(x))
As x goes to infinity, 1/2 becomes useless, and for simplicity sake let's remove it.
1/x < 1/(2sqrt(x)) --> 1/x < 1/sqrt(x)
Then we can clearly see that as x increase to infinity, the rate of change of sqrt(x) is greater then ln(x).
I've just though about it while reading your question, but I am fairly sure of my answer.
Vardd
This is the way to do it. You also need to show that it's a positive value from the beginning though (lim x approaches 0 of f(x) = +) since increasing wouldn't really prove anything by itself (ex: any function where f(x) = - where x< some positive number k, but f(x) increasing from x>0).JonF said:Showing the derivative of x^1/2 – ln(x) is positive for all x>0 is fairly easy.
Anonymous217 said:This is the way to do it. You also need to show that it's a positive value from the beginning though (lim x approaches 0 of f(x) = +) since increasing wouldn't really prove anything by itself (ex: any function where f(x) = - where x< some positive number k, but f(x) increasing from x>0).
grossgermany said:Except the statement is not true
is not always positive for x>0.
fourier jr said:whoops, it's not monotone like I said but the minimum is positive, & that will still do it
alomari2010 said:consider the function [tex]{f(x) = e^{\sqrt{x}} - x,}[/tex] which is >0
so that [tex]f^{\prime}[/tex] >0 for all x and therefore
[tex]e^{\sqrt{x}} - x > 0[/tex] , and thus [tex]\sqrt{x} > \ln{x}[/tex]
To begin, you can start by assuming that x is greater than 0. Then, you can use the definition of ln(x) and the properties of logarithms to manipulate the inequality into a form that is easier to prove. From there, you can use algebraic techniques and properties of exponents to simplify and eventually prove the inequality.
Some key properties include the power rule, product rule, quotient rule, and change of base rule. These properties allow you to manipulate the logarithmic expression into a form that is easier to work with and can help prove the inequality in question.
Yes, you can use calculus to prove this inequality. One approach could be to take the derivative of both sides of the inequality with respect to x and show that the resulting expression is always positive, thus proving that the original inequality holds for all x>0.
Yes, it may be helpful to consider both small and large values of x when proving this inequality. For small values of x, you can use the limit definition of ln(x) to show that it approaches 0, while for large values of x, you can use the limit definition of sqrt(x) to show that it approaches infinity.
You can use mathematical induction to prove this inequality by first showing that it holds for a base case (e.g. x=1) and then assuming that it holds for some arbitrary value of x (the induction hypothesis). From there, you can use algebraic techniques and the properties of logarithms to show that the inequality holds for x+1, thus proving the statement for all x>0.