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I have been trying to figure out the solution to a logarithm problem, and keep running into the equation $\\10\log^2_2(x) = x$. In so doing I've been trying to simplify $\log_2^2(x)$. What I have is

$\log_2^2(x) = \log_2(x)\log_2(x) = \log_2(x^{\log_2(x)})$

which as far as I can tell is no easier to work with.

Am I missing something? Or is there a completely different alternative way to look at this problem? Thanks!

saz
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Iinferno1
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    Except for some fine-tuned cases, equations involving both polynomials and logarithms are not solvable in terms of elementary functions. But they are sometimes solvable if you introduce a new function called Lambert W function. For instance, your equation has 3 real zeros (which are all positive), and they can be written in terms of this function. – Sangchul Lee Oct 09 '16 at 05:50

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This is a fun one! We can solve it using the Lambert W Function, the inverse of the function

$$f(x) = xe^x$$

$$10\log_2^2(x) = x$$

$$\sqrt{10}\log_2(x) = \sqrt{x}$$

$$2^{\log_2(x)\sqrt{10}} = 2^{\sqrt{x}}$$

$$x^{\sqrt{10}} = 2^{\sqrt{x}}$$

$$x = 2^{\frac{\sqrt{x}}{\sqrt{10}}}$$

$$x = e^{\frac{\sqrt{x}}{\sqrt{10}}\log 2}$$

$$(\sqrt{x})^2 = e^{\frac{\log 2}{\sqrt{10}}\sqrt{x}}$$

$$\sqrt{x} = e^{\frac{\log 2}{2\sqrt{10}}\sqrt{x}}$$

Let $c = \frac{\log 2}{2\sqrt{10}}$:

$$\sqrt{x} = e^{c\sqrt{x}}$$

$$-c\sqrt{x}e^{-c\sqrt{x}} = -c$$

$$-c\sqrt{x} = W(-c)$$

$$\sqrt{x} = \frac{W(-c)}{-c}$$

$$x = \frac{(W(-c))^2}{c^2}$$

$$x = \frac{\left(W\left(-\frac{\log 2}{2\sqrt{10}}\right)\right)^2}{\left(\frac{\log 2}{2\sqrt{10}}\right)^2}$$

$$x =\left(W\left(-\frac{\log 2}{2\sqrt{10}}\right)\right)^2\frac{\log^2 2}{40}$$

There are actually $3$ solutions for $x$. One results in taking this function as is. The other results in using $-\sqrt{10}$ instead of $\sqrt{10}$. The last results in taking $\sqrt{10}$ and using the second branch $W_{-1}$ of the Lambert W function (since there are multiple solutions to $xe^x = y$ for all $y<0$).

Carl Schildkraut
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  • There's one solution around 0.8 and a second at around 1.2. The last is around 990, and uses the other branch $W_{-1}$, of the Lambert W. – Parcly Taxel Oct 09 '16 at 05:59
  • Plotting these functions, I see that now. What did I do wrong? – Carl Schildkraut Oct 09 '16 at 05:59
  • You haven't done anything wrong. $\ddot\smile$ I was just using WA to check the numerical solutions. – Parcly Taxel Oct 09 '16 at 06:01
  • @ParclyTaxel Can you evaluate $W$ numerically with WA? I tried and was unable to verify that my answer is indeed correct. Also, why did I miss 2 of the 3 solutions? (I get why I missed the second branch, but why are there 3? – Carl Schildkraut Oct 09 '16 at 06:02
  • The second real zero comes from by choosing different sign for the square root, i.e., $$\sqrt{10}\log_2(x) = -\sqrt{x}.$$ The final zero comes from the same equation $\sqrt{10}\log_2(x) = \sqrt{x}$ but with different branch cut of $W$. – Sangchul Lee Oct 09 '16 at 06:10
  • @SangchulLee Thanks for clearing that up! – Carl Schildkraut Oct 09 '16 at 06:13
  • ProductLog[x] or LambertW[x] will evaluate the Lambert W in Wolfram Alpha. Using ProductLog[-1, x] (same for LambertW) gives the other branch. – Parcly Taxel Oct 09 '16 at 06:16