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Hello everyone what is bigger $\frac{3}{e}$ or $\ln(3)$?

I tried to square it at $e$ up and I got:

$e^{\frac{3}{e}} = \left(e^{e^{-1}}\right)^{3\:}$ and $3$ but I don't know how to continue I also tried to convert it to a function but I didn't find.

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  • $\begingroup$ Your problem is equivalent to determining whether $\ln 3 - \ln (\ln 3)$ or $1$ is greater. If you let $f(x) = \ln x - \ln (\ln x)$, then you can find the minimum using calculus and find the number which is larger. $\endgroup$
    – Toby Mak
    Commented Jun 29, 2020 at 13:35

4 Answers 4

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Note that $$\dfrac d{dx}\left(\dfrac{x}{\ln x}\right)=\dfrac{\ln x-1}{(\ln x)^2}$$ Therefore, this function takes minimum value at $e$. Hence, $$\dfrac{3}{\ln3}>\dfrac{e}{1}\\ \implies\boxed{\dfrac 3e>\ln3}$$

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Define $f(x)= {x \over e}-ln(x)$. Note that $$f'(x)= {1\over e}-{1\over x}$$

So $f'(x)\gt 0 $ for $x \gt e$. Thus $f(x)$ is increasing for $x \gt e$

Now, note that $f(e)=0$ and $3\gt e$.

Thus $f(3) \gt 0$ as $f(e)=0$ and $f(3) \gt f(e)$

So $f(3)= {3 \over e}-ln(3) \gt 0$

Thus $$ {3 \over e}\gt ln(3) $$

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  • $\begingroup$ Infact you can use this to show $f(x) \gt 0$ for all $x \gt 0$. $\endgroup$ Commented Jun 29, 2020 at 14:22
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Let's assume $f(x)=\frac{x}{e}$ and $g(x)=\ln(x)$.

$f(x)=g(x)$ for $x=e$ and $f$ grows faster than $g$. So I would say

$\frac{3}{e}>\ln(3)$, since $3>e$.

I hope this is what you are looking for.

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Hint:

Compare the logs and use that $\ln $ is concave, hence its representative curve is below each of its tangents.

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