Comparing $\ln 1000$, $\sqrt[5]{1000}$, $3^{1000}$, and $1000^{15}$ without calculator

Question

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!! :)

Answer 1

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

$$ a^b = c^{b\log_c(a)} $$

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

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

and

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

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

All you need to do is show that $1000\log_{10}(3) > 45$. 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 $\frac{3}{2} < 10^{1/5}$).

Answer 2

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

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

Answer 3

Without a calculator you may proceed as follows:

• $4^5 = 1024 \Rightarrow \sqrt[5]{1000} < 4$
• $2 < e < 3 \mbox{ and } 2^{10} > 1000 = e^{\ln{1000}} > 3^6 \Rightarrow 6 < \ln 1000 < 10$
• $3^{1000} = \left(3^{\frac{1000}{15}}\right)^{15} > \left(3^{60}\right)^{15} > 1000^{15}$

Answer 4

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 $$\sqrt[5]{1000}<\ln 1000<1000^{15}<3^{1000}.$$