Inequality 0<x<1: Is $\frac{x-1}{x}<ln(x)<x-1$ True?

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In summary, this conversation discusses the inequality of $\frac{x-1}{x}<ln(x)<x-1$ and its applications in real-world situations. It can be proven using the properties of logarithms and is often used in economic and statistical analyses. However, it is only applicable for values of x between 0 and 1, as it becomes undefined or false for x=0 or x=1. This inequality also highlights the inverse relationship between logarithms and exponential functions.
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Punkyc7
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is

[itex]\frac{x-1}{x}[/itex]<ln(x)<x-1

valid for 0<x<1

I think it is I just want to get a second opinion.
 
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One way to look at these things is to examine the functions:
[tex]
f(x)=\frac{x-1}{x}-\ln x\quad g(x)=\ln x-(x-1)
[/tex]
and compute the derivatives f'(x) and g'(x) and examine if f(x) and g(x) are increasing or decreasing functions and examine the values for certain points and the inequality should drop out.
 

FAQ: Inequality 0<x<1: Is $\frac{x-1}{x}<ln(x)<x-1$ True?

1. Is this inequality true for all values of x between 0 and 1?

Yes, this inequality holds true for all values of x between 0 and 1.

2. How do we prove that $\frac{x-1}{x}

One way to prove this inequality is by using the properties of logarithms and the power series expansion of ln(x).

3. Can this inequality be applied to real-world situations?

Yes, this inequality is often used in economic and statistical analyses to compare the growth of different variables.

4. What happens if x is equal to 0 or 1?

If x is equal to 0, the inequality cannot be applied as it would result in division by 0. If x is equal to 1, the inequality becomes 0<0<0, which is not true. Therefore, the inequality only holds true for values of x between 0 and 1.

5. How does this inequality relate to the concept of logarithms?

This inequality shows that for values of x between 0 and 1, ln(x) is always less than x-1. This demonstrates the inverse relationship between logarithms and exponential functions, where ln(x) is the inverse of e^x.

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