Krdytkyu

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!! :)

Without a calculator you may proceed as follows:

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


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


• $3^{1000} = left(3^{frac{1000}{15}}right)^{15} > left(3^{60}right)^{15} > 1000^{15}$

$
ln(1000)= 3ln(10)
$ is between 6 and 9 because $2<ln(10)<3$

this is clearly bigger than $1000^{0.2}$ because $$ (ln(1000))^5 > 6^5>4^5=1024>1000$$

You can use a change of base on the exponentials to see which should be larger:

$$
a^b = c^{blog_c(a)}
$$

We can use base 10 or, as I prefer, the natural log:

$$
3^{1000} = e^{1000ln(3)} approx e^{1099}
$$

and

$$
1000^{15} = e^{15ln(1000)} approx e^{104}
$$

It should be clear that the $5^text{th}$ root of $1000$ will be smaller, but you can proceed the same:

$$
sqrt[5]{1000} = 1000^frac{1}{5} = e^{frac{1}{5}ln(1000)} approx e^{1.4}
$$

EDIT:

I just realized you were supposed to not use a calculator. Because of this, the base $10$ logarithm makes more sense:

begin{align*}
3^{1000} &= 10^{1000*log_{10}(3)} \
1000^{15} &= 10^{15*log_{10}(1000)} = 10^{45} &text{$log_{10}left(10^3right) = 3$}\
1000^{frac{1}{5}} &= 10^{frac{1}{5}log_{10}(1000)} = 10^{frac{3}{5}} &text{$log_{10}left(10^3right) = 3$}
end{align*}

All you need to do is show that $1000log_{10}(3) > 45$. We can approximate $log_{10}(3)$. We know that $3^2 < 10 < 3^3$, therefore $10^frac{1}{3} < 3 < 10^frac{1}{2}$ and thus $frac{1}{3} < log_{10}(3) < frac{1}{2}$. We use the smaller number (we want to "over" correct, by going as small as possible) and find that:

$$
frac{1000}{3} > 45
$$

Second Edit : $ln(1000)$

Again, let's switch to base 10:

$$
log_a(x) = frac{log_b(x)}{log_b(a)}
$$

So we have:

$$
ln(1000) = frac{log_{10}(1000)}{log_{10}(e)} = frac{3}{log_{10}(e)}
$$

We can approximate $log_{10}(e)$ in the same way we approximated $log_{10}(3)$. We know that $2 < e < 3$. $2^3 < 10 < 2^4$ and (again) $3^2 < 10 < 3^3$, therefore $frac{1}{4} < log_{10}(2) < frac{1}{3}$ or $frac{1}{3} < log_{10}(3) < frac{1}{2}$. Be careful here, this will tell us that:

begin{align*}
frac{3}{frac{1}{2}} < ln(1000) < frac{3}{frac{1}{4}} \
6 < ln(1000) < 12
end{align*}

If you can show that $10^frac{3}{5}$ is less than $6$ or greater than $12$, then you have your answer. So, surely $10^{frac{1}{5}} < 2$ (since $2^5 = 32$) but also $2^3= 8 > 6$ (so that's no good for our range). This is tricky to do, but can be done with some trial and error (hint: $1.5 = frac{3}{2}$ won't work--you'll end up proving that $frac{3}{2} < 10^{1/5}$).

From $$e^2<3^2<10<e^3<3^3$$ we have $$2<2ln 3<ln 10<3<3ln 3 tag 1$$

• 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 $(1):$

$$ln3<ln6< color{blue}{ln(ln 1000)}<ln9=2ln3 $$
$$color{blue}{ln3^{1000}}=1000ln3$$
$$90ln3<color{blue}{ln 1000^{15}}<135ln3$$
The numbers are in the same order as their natural logarithmes.

Putting together both parts, we conclude $$sqrt[5]{1000}<ln 1000<1000^{15}<3^{1000}.$$