How this be solved by inequalities? I tried it 3 days

Question

There are two positive integers $n\ge3$ for which the expression $$\frac{\log2×\log3×\log4×\cdots×\log(n-1)}{10^n}$$ takes on its smallest value. What is the larger of these two integers?

I tried am gm inequalities and I tried to do it by calculus by maximal minima differntiation method and I don't know how to differentiate this type of function

Answer 1

Let $$f(n)=\frac{\log2\log3\cdots\log(n-1)}{10^n}$$ where $\log=\log_{10}$. Comparing the form of $f(n)$ to $f(n+1)$ shows that $f(n+1)=f(n)\frac{\log n}{10}$, and this tells everything about $f$'s minimum value:

• For $n<10^{10}$ the multiplier is less than $1$, so $f(n)>f(n+1)$;
• For $n=10^{10}$ the multiplier equals $1$, so $f(n)=f(n+1)$;
• For $n>10^{10}$ the multiplier is more than $1$, so $f(n)<f(n+1)$.

So the minimum is achieved by $f(10^{10})$ and $f(10^{10}+1)$ and the requisite integer is $10^{10}+1$.

Answer 2

Well, use the concept of calculus.

The smallest value of $f(n)$ would be when $f(n)$ stops getting smaller and starts getting larger.

So we want to compare $f(n+1)$ to $f(n)$ to find out whe $f(n+1) < f(n)$ stops being true and $f(n+1) > f(n)$ starts being true.

So.... solve. If we let $??$ be either $=, <,$ or $>$ then:

$f(n+1) = \frac{\log 2\log 3....\log(n-1)\log n}{10^{n+1}} ??\frac{\log 2\log 3....\log(n-1)}{10^n}=f(n) \implies$

$\frac{\log 2\log 3....\log(n-1)\log n}{\log 2\log 3....\log(n-1)} ?? \frac {10^{n+1}}{10}\implies$

$\log n ?? 10\implies$

$10^{\log n} = n ?? 10^{10}$

So if $n < 10^{10}$ we have $f(n)$ is getting smaller (and $n < 10^{10} \iff f(n+1) < f(n)$). And if $n = 10^{10}$ we have $f(n+1) = f(n)$ (but that is the only such point). And at $n > 10^{10}$ we have $f(n)$ is getting bigger and $f(n+1) > f(n)$.

So $f(n)$ is at it's smallest at $n=10^{10}$ (when $f(n)$ stops getting smaller) and at $n=10^{10} + 1$ (when $f(n)$ starts getting larger. The larger number where $f(n)$ is smallest is $10^{10} + 1$ (and the smaller is $10^{10}$).