How to compare log and exponential

Question

$x^{1/3}$

$　 (\log_{2}{x})^3$

How to compare them?

Answer 1

As others have mentioned, there's no closed-form solution to $$x^{1/3} = (\log_2 x)^3$$ in elementary functions, but it can be solved using the Lambert W function.

Briefly, the Lambert W function is the (multi-branched) inverse function of $$f(x) = xe^x$$ That is, $$x = W_k(x)e^{W_k(x)}$$ When working with real numbers, we only need the $W_{-1}$ and $W_0$ branches.

This function is generally not found in the standard math libraries of common programming languages, but it is found in most advanced mathematics packages. You can approximate it in various ways, eg, Newton's method.

Anyway, here's how to find the (real) solutions to your equation, assuming that you have some way of evaluating the Lambert W.

$$x^{1/3} = (\log_2 x)^3$$ $$x^{1/9} = \log_2 x$$ Let $x=u^9$ $$u = \log_2 u^9 = 9\log_2 u$$ $$2^u = u^9$$ $$u\log 2 = 9\log u$$ $$u = 9\log u / \log 2$$

The $\log u$ function grows slower than any power $u^n$ for $n>1$, so as Paul Rebenciuc mentioned at the end of his answer, for all sufficiently large $u$, $2^u > u^9$.

Let's pause for a moment to look at the graph of $y = 9\log x / \log 2$ and see where it crosses $y=x$.

Here are close-ups of the cross-over points.

So we have solutions around $u = 1.087$ and $u = 51.069$.

Continuing,

$$u = 9\log u / \log 2$$ Let $m = 9 / \log 2$ $$u = m \log u$$ Let $u = e^{-v}$, so $\log u = -v$ $$e^{-v} = -mv$$ $$-1/m = ve^v$$ Now we can apply the Lambert W: $$W(-1/m) = W(ve^v) = v$$

So now we can get $u$ from $u = e^{-v}$ and $x$ from $x=u^9$. As I mentioned earlier, we need to evaluate both the $W_{-1}$ and $W_0$ branches to get both solutions.

Here's a short Sage / Python script that performs this calculation. It also verifies that $u = -mv$.

Code

m = 9 / log(2)
for v in [lambert_w(i, -1/m) for i in (-1, 0)]:
    u = exp(-v)
    print("v", v.n())
    print("u", u.n(), "-mv", (-m*v).n())
    print("u^9", (u^9).n(), "2^u", (2^u).n())
    print()

Output

v -3.93318771959389
u 51.0695137614913 -mv 51.0695137614913
u^9 2.36295530009418e15 2^u 2.36295530009419e15

v -0.0837437490868883
u 1.08735022361793 -mv 1.08735022361793
u^9 2.12483412771106 2^u 2.12483412771106

And here's a link containing the script which runs on the SageMath server.

Answer 2

Have you tried this? If not, take a look. Yes, there is no method to exactly compare them, unless you consider this special function $W_z(k)$, as a known function.

https://www.wolframalpha.com/input/?i=x%5E%281%2F3%29-log2%28x%29%5E3

If you refer to the behavior of log/power to infinity (i.e. for x sufficiently large), always the power $x^n, n>0$ goes faster to $\infty$ than $(log_mx)^p$, where $p,m>1$.

Answer 3

Cubing, consider $$f(x)=x-\frac{\log ^9(x)}{\log ^9(2)}$$ $$f'(x)=1-\frac{9 \log ^8(x)}{x \log ^9(2)}$$ $$f''(x)=\frac{9 (\log (x)-8) \log ^7(x)}{x^2 \log ^9(2)}$$ The first derivative cancels at two points $$x_{1,2}=\exp\Big[-8 W(\pm k) \Big]\qquad \qquad k=\frac{\log ^{\frac{9}{8}}(2)}{8 \sqrt[4]{3}}$$ where $W(.)$ is Lambert function (numerically, $x_1=0.622419$ and $x_2=1.71274$). $$f(x_1) \sim 0.655209 \qquad \qquad f''(x_1)=+28.7147$$ $$f(x_2) \sim 1.610341 \qquad \qquad f''(x_2)=-8.09651$$

So, $x_1$ corresponds to a minimum and $x_2$ to a maximum.

Now, by inspection, $f(2)=1$. So, using Newton method, the iterates will be $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.00000 \\ 1 & 2.18208 \\ 2 & 2.13183 \\ 3 & 2.12495 \\ 4 & 2.12483 \end{array} \right)$$ Then $f(x) >0$ for any $x < 2.12483$