Is ln(2) Greater Than (2/5)^(2/5)?

  • MHB
  • Thread starter anemone
  • Start date
In summary, "Prove ln 2 > (2/5)^2/5" is a mathematical statement that asks to prove that the natural logarithm of 2 is greater than the fifth root of the fraction 2/5. This statement is important because it is a fundamental inequality in mathematics with various applications in fields such as calculus, probability, and statistics. It can be proved using methods such as the properties of logarithms, calculus, or mathematical induction. This inequality also has real-life applications in finance, economics, physics, and biology.
  • #1
anemone
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MHB
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Here is this week's POTW:

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Prove $\ln 2>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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  • #2
Hi MHB,

I will give the community another week to attempt at last week's high school POTW. I welcome anyone of you who are interested in this problem to give it one more try and I am looking forward to receiving your submission!(Happy)
 
  • #3
Solution from other:

Summing just the $k=0$ and $k=1$ terms from the identity \(\displaystyle \ln 2=\sum_{k=0}^\infty \dfrac{2}{2k+1}\left(\dfrac{7}{31^{2k+1}}+\dfrac{3}{161^{2k+1}}+\dfrac{5}{49^{2k+1}}\right)\) gives

$\ln 2>\dfrac{29558488681560}{42643891494953}\\ \ln2>0.693147\\ (\ln2)^5>(0.693147)^5 \\ (\ln 2)^5>0.160002 \\ (\ln 2)^5>\dfrac{4}{25}\\ \ln2>\left(\dfrac{2}{5}\right)^{\frac{2}{5}}$
 

FAQ: Is ln(2) Greater Than (2/5)^(2/5)?

What is the significance of proving ln 2 > (2/5)^2/5 in POTW #400?

The problem of the week (#400) is a challenge presented to individuals to solve a mathematical problem. Proving ln 2 > (2/5)^2/5 is the main objective of this specific problem.

How do you approach solving this problem?

One approach to solving this problem is by using the properties of logarithms and exponents. You can also use the fact that ln 2 is approximately equal to 0.693 and (2/5)^2/5 is approximately equal to 0.729.

Can you provide a step-by-step solution to this problem?

Step 1: Rewrite (2/5)^2/5 as (2/5)^4/5 to make it easier to compare with ln 2.

Step 2: Use the change of base formula to rewrite ln 2 as log2/log(e).

Step 3: Compare log2/log(e) with (2/5)^4/5. You will notice that (2/5)^4/5 is a larger value than log2/log(e).

Step 4: Since (2/5)^4/5 is larger than log2/log(e), we can conclude that ln 2 is also larger than (2/5)^2/5.

Are there any other methods to prove ln 2 > (2/5)^2/5?

Yes, there are other methods to prove this inequality. One method is by using calculus and finding the derivative of both ln 2 and (2/5)^2/5, and then comparing their values at certain points. Another method is by using the Taylor series expansion of ln x and approximating the values of ln 2 and (2/5)^2/5.

What are the applications of this problem in real life?

The concept of logarithms and exponents is widely used in various fields such as finance, science, and engineering. This problem, in particular, can be applied in calculating compound interest, population growth, and radioactive decay. It also helps in understanding the relationship between different exponential functions and their growth rates.

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