Which function is eventually larger: $\ln(x)\text{ or }\sqrt[3]x$?

Question

So I was scrolling through the homepage of Youtube when I came across this video by Blackpenredpen asking which of the following functions was bigger:$$\ln(x)\text{ or }\sqrt[3]x$$which I thought that I might be able to figure out. Here is my attempt at finding out which of the functions was eventually bigger:$$\ln(x)=\sqrt[3]x$$$$x=e^{x^\frac{1}{3}}$$$$x\approx6.4057$$Now knowing this, now all we need to do is take a number that is greater than what we got for the value of $x$ and see which of the following functions returns a greater number:$$\ln(7)\quad\overset{?}{\large\circ}\quad\sqrt[3]7$$$$\ln(7)\approx1.94591014906$$$$\sqrt[3]7\approx1.91293118277$$$$\text{Therefore, the natural logarithm is eventually larger than the cube root function}$$

My question

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Is what I have a sufficient proof that the natural logarithm is eventually bigger than the cube root function, or are there any flaws in my proof?

Answer 1

You can study the variations of the function $f(x)=\ln(x)-x^\frac13$. You have $f'(x) = \frac1x-\frac1{3x^\frac23}$. Solving for $f'(x)=0$ gives the solution $x=27$.

Therefore $f'(x) \geq 0$ if $x \in ]0,27]$ and $f'(x) \leq 0$ if $x \geq 27$. Also, note that $f(x) \to_{x \to 0} -\infty$, $f(27)\approx 0.296 > 0$ and $f(x) \to_{x \to +\infty} -\infty$. So first $f$ is increasing from$ -\infty$ to $\approx0.296$ and then $f$ is decreasing from $\approx0.296$ to $-\infty$.

This means there are two critical values (since $f$ is continuous) such that $f(x)=0$, approximated by $x_1 \approx 6.164$ and $x_2 \approx 93.354$ (there is no analytical expression without using special functions) such that $\ln(x) \geq x^\frac13$ for $x \in [x_1,x_2]$. Otherwise $\ln(x) \leq x^\frac13$ for $x \in ]0,x_1]\cup[x_2,+\infty[$.

For instance you can see that $\ln(100) \approx 4.605$ while $100^\frac13 \approx 4.6415 > \ln(100)$ so your conclusion is incorrect.

Answer 2

For any fixed $a$ and $b$, where $a>0, b>0$, you can show, for a sufficiently large $x$, you have

$$x^a>\ln^b x$$

Since $a=b\cdot\frac{a}b$, so

$$x^a>\ln^b x\Leftrightarrow x^{a/b}>\ln x=\frac{b}{a}\cdot\ln x^{a/b}$$

Let $t=x^{a/b}$, so it is equivalent to show

$$\frac{a}b \cdot t>\ln t$$

Since $\frac{a}b>0$, it is true for sufficiently large $t$, hence, hold for sufficiently large $x$.