# Getting stuck on simple logarithmic equation

$$x \times \ln (x) = 1$$

I am trying to solve that equation. I used the theory $ln(a) = ln(b)$ being equivalent to $a = b$ and got stuck at

$$x = e^{\frac{1}{x}}$$

That's as far as I went and I know there's a solution (around 1.8 or 1.9), since I used my calculator, but I'd like to know how to do this by hand.

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You'll probably have to use the Lambert W function. –  Alex Becker Apr 8 '13 at 21:32
This is not an equation that can be solved with elementary functions, as far as I can tell. –  Sammy Black Apr 8 '13 at 21:34

You can use the law of logarithms which states that for $a,b\in\mathbb{R}$: $a\ln{b}=\ln\left(b^{a}\right)$.

Therefore, you have:

$$x\ln{x}=1 \implies \ln{x^{x}}=1$$

You hence have:

$$x^{x}=e$$

Which does not have an elementary closed form, so you must use numerical methods (for instance Newton-Raphson iteration) to get an approximation (Mathematica gives $x\approx 1.76322$).

If you're interested, the closed form solution is: $$\frac{1}{W(1)}, \qquad \text{ where } W(z) \text{ is the LambertW function}$$

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There is no solution using only algebraic manipulation. We have to use the "product-log" or Lambert-W function to solve this, and this function doesn't fall in the "simple functions" category. :)

Basically, the Lambert-W function is the inverse function of: $$f(x) = xe^x$$ Equivalently: $$x=W(xe^x)$$

So, using your expression: $$x=e^\frac{1}{x}$$ $$1=\frac{1}{x}e^\frac{1}{x}$$ Taking the product-log of both sides: $$W(1)=W\left(\frac{1}{x}e^\frac{1}{x}\right)$$ $$W(1)=\frac{1}{x}$$ $$x=\frac{1}{W(1)}$$

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This equation won't have an elementary solution - you'll have to solve it numerically. (Or you can ask WA).

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It's in fact impossible to express your variable $x$ as a combination of usual elementary function. Symbolically, you shall express it using Lambert's L function, which solves the equation

$$x=L(x)e^{L(x)}$$

For instance, if $$xe^x = 1$$ then $x=L(1)$ as is easily seen. By expressing $x$ as a the logarithm of some other number $y$, one has from the preceding equation

$$y\ln y=1$$

Then, the solution of your equation would be $\ln(y)=L(1)$ or $y=e^{L(1)}$. Take a look at wikipedia's http://en.wikipedia.org/wiki/Lambert_W_function for further information!

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$L{}$ function? –  apnorton Apr 8 '13 at 21:52

${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$See Example 4 here: http://en.wikipedia.org/wiki/Lambert_W_function

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(Not my downvote.) The community typically likes answers that are self-contained, rather than just links to outside websites. This might work well as a comment, though. –  apnorton Apr 8 '13 at 21:38
@anorton Thanks for letting me know. I think that the criterion does not make sense (even though I will respect that), especially if that outside website is Wikipedia, and the link contains the answer + other things that the OP may want to know about the problem he/she is dealing with. The "downvote and go," that is what I do not understand (not that I care about my "score" here). –  Lord Soth Apr 8 '13 at 21:53
I understand. I also dislike it when someone just downvotes one of my answers without any feedback whatsoever. I think the reasoning for self-contained answers is that links can go bad over time (for example, someone could renumber the examples on Wikipedia), but answers here should be time-independent. –  apnorton Apr 8 '13 at 21:55