5

I want to prove that $3^{100} > 5\cdot10^{47}$ without using calculator or any approximation of the logarithm. I tought about finding an integral representation of $(3^{100} - 5\cdot10^{47})$ with the integrand non negative in the chosen integration interval (something like this), but I don't know how to chose the integrand. Do you have any ideas? Or even any elegant way to prove the inequality?

J. W. Tanner
  • 60,406
user967210
  • 711
  • Maybe relevant: https://math.stackexchange.com/q/2276227/25554 – MJD Feb 13 '24 at 17:25
  • 1
    “Without any approximation of the logarithm” is pretty broad. For example, is it really forbidden to argue that $3^5< 2^8 < 3^6$ therefore $\frac58<\log_3{2} < \frac 68$? – MJD Feb 13 '24 at 19:23
  • @MJD no it's not forbidden! I just wanted to know if there is an alternative way, and maybe use that in other kind of problems like that. If you have a solution, please feel free to post, thank you! – user967210 Feb 13 '24 at 19:40
  • If logarithms base $~10~$ are allowed, then the problem becomes trivial because $~\log_{10}(3) \approx 0.477,~$ while $~\log_{10}(5) \approx 0.699.$ – user2661923 Feb 13 '24 at 19:59
  • 3
    @user2661923 But then the question is where you can obtain such good approximations of $\log_{10} 3$ and $\log_{10} 5$ without extensive floating-point calculations. The two sides are quite close ($47.71 > 47.70$) so tricks like the one I suggested, based on small integers and inaccurate estimates, will not suffice. – MJD Feb 13 '24 at 20:33
  • @MJD No calculator needed. See this answer. – user2661923 Feb 13 '24 at 21:37

1 Answers1

7

We have by hand calculation

  • $3^{10} =59049>59\cdot 10^3$
  • $59^2=3481>348\cdot10$
  • $348^2=121104>12\cdot 10^4$

then

$$3^{100}>348\cdot 12^2\cdot 10^{43}>5.01\cdot 10^{47}$$

user
  • 154,566
  • 2
    I'd say given that the bound is tight, this has to be the most efficient way to solve by approximation(+1) – D S Feb 13 '24 at 21:14
  • @DS Yes I think it is a reasonable way using only hand calculation. – user Feb 13 '24 at 21:16
  • @DS We can improve a little bit using $348^2>12\cdot 10^4$. – user Feb 13 '24 at 21:23
  • @user very smart and simple! I know that the integral method would be much more complex, but do you think that it's possible to find a suitable integrand function? – user967210 Feb 13 '24 at 21:32
  • 1
    @user967210 I'm not sure how can find such kind of solution, maybe someone else is working on that. Glad you appreciate also this alternative way. Bye – user Feb 13 '24 at 21:35
  • Unclear if the logarithm approach in an answer that I linked to, in a comment following the original question, is good enough to confidently solve the problem. The approach definitely gives 3 decimal places for $~\log_{10} (2)~$ and $~\log_{10} (3).~$ This becomes borderline sufficient. Going for the 4th decimal place depends on getting the 4th decimal place for $~\log_{e}(10).~$ Unclear if that kind of accuracy can be confidently attained, using my approach. – user2661923 Feb 13 '24 at 21:45