Comparing $\ln 1000$, $\sqrt[5]{1000}$, $3^{1000}$, and $1000^{15}$ without calculator In my Pre-Calculus class we were given the following problem:

Put the following four values in order from smallest to largest: $\ln 1000$, principal $5$th root of $1000$, $3^{1000}$, and $1000^{15}$.

Using a calculator, I was able to figure out that the correct order is as follows: principal $5$th root of $1000, \ln 1000, 1000^{15}, 3^{1000}$. 
However, we are supposed to be able to solve this problem without a calculator, and then it becomes much more difficult for me, to say the least. 
I know that any exponential growth function (e.g., $3^x$) "eventually" outpaces any polynomial function (e.g., $x^{15}$). 
So I suppose the teacher could just be assuming that we will see $x = 1000$ and conclude that $1000$ seems "sufficiently large" for the exponential, $3^{1000}$, to have outpaced $1000^{15}$. Nonetheless, that is somewhat unsatisfying.
Furthermore, it still provides no answer to the even trickier question of how you know $\ln 1000$ is greater than the principal $5$th root of $1000$. This becomes clear when you graph on a calculator, but as I said I'm supposed to be able to solve this problem without a calculator. And, indeed, the difference between $\ln 1000$ and the principal $5$th root of $1000$ is only about $3$, according to my calculator -- so which of the two is bigger seems hardly obvious to me from a quick glance at the two quantities (without plugging them into a calculator). 
It's an incredibly fascinating problem, but I just can't figure it out. Any answers would be greatly appreciated. Thank you so much!! :) 
 A: Without a calculator you may proceed as follows:


*

*$4^5 = 1024 \Rightarrow \sqrt[5]{1000} < 4$

*$2 < e < 3 \mbox{ and } 2^{10} > 1000 = e^{\ln{1000}} > 3^6 \Rightarrow 6 < \ln 1000 < 10$

*$3^{1000} = \left(3^{\frac{1000}{15}}\right)^{15} > \left(3^{60}\right)^{15} > 1000^{15}$
A: $
\ln(1000)= 3\ln(10) 
$ is between 6 and 9 because $2<\ln(10)<3$
this is clearly bigger than $1000^{0.2}$ because $$ (\ln(1000))^5 > 6^5>4^5=1024>1000$$
A: From $$e^2<3^2<10<e^3<3^3$$ we have $$2<2\ln 3<\ln 10<3<3\ln 3 \tag 1$$


*

*Realize that $\sqrt[5]{1000}<\ln1000,$ as their fifth powers are in this order: $$1000<2^{10}<6^5<(\ln1000)^5$$

*Now, compute logarithmes of $\ln 1000,\;1000^{15},\;3^{1000}$ and find their estimates with the use of $(1):$
$$\ln3<\ln6< \color{blue}{\ln(\ln 1000)}<\ln9=2\ln3 $$
$$\color{blue}{\ln3^{1000}}=1000\ln3$$
$$90\ln3<\color{blue}{\ln 1000^{15}}<135\ln3$$
The numbers are in the same order as their natural logarithmes. 
Putting together both parts, we conclude $$\sqrt[5]{1000}<\ln 1000<1000^{15}<3^{1000}.$$ 
