What Is Bigger $\frac{3}{e}$ or $\ln(3)$

Question

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.

Answer 1

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, $$\dfrac{3}{\ln3}>\dfrac{e}{1}\\ \implies\boxed{\dfrac 3e>\ln3}$$

Answer 2

Hint:

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

Answer 3

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.

Answer 4

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