# What Is Bigger $\frac{3}{e}$ or $\ln(3)$ [duplicate]

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.

• 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. Commented Jun 29, 2020 at 13:35

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

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

• Infact you can use this to show $f(x) \gt 0$ for all $x \gt 0$. Commented Jun 29, 2020 at 14:22

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.

Hint:

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