# 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!! :)

• Welcome to MathSE. Please read this tutorial on how to typeset mathematics on this site. – N. F. Taussig Dec 2 '18 at 12:26

Without a calculator you may proceed as follows:

• $$4^5 = 1024 \Rightarrow \sqrt[5]{1000} < 4$$
• $$2 < e < 3 \mbox{ and } 2^{10} > 1000 = e^{\ln{1000}} > 3^6 \Rightarrow 6 < \ln 1000 < 10$$
• $$3^{1000} = \left(3^{\frac{1000}{15}}\right)^{15} > \left(3^{60}\right)^{15} > 1000^{15}$$
• Or for the last without fractions in the exponent, $3^{1000}=(3^{20})^{50}>10^{50}>10^{45}=1000^{15}$ – Lutz Lehmann Dec 2 '18 at 10:49

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

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

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 $$\frac{3}{2} < 10^{1/5}$$).

• Wait am I supposed to be able to estimate stuff like 1000*ln(3) without a calculator? – Will Dec 2 '18 at 5:47
• @Will Yeah, I just realized that. I edited it. It should make more sense now. – Jared Dec 2 '18 at 5:56
• I will read over it. I'm guessing it will make much more sense. Thank you so much!! One thing: Do you have any idea how to prove that ln 1000 is larger than the root but smaller than either the exponential or the power? You've proven three of the four numbers in the original problem, and I am so thankful. But do you have any thoughts on the other one, ln 1000? – Will Dec 2 '18 at 5:58

From $$e^2<3^2<10 we have $$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 $$\sqrt[5]{1000}<\ln 1000<1000^{15}<3^{1000}.$$