Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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.

share|improve this question
    
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
add comment

5 Answers

up vote 6 down vote accepted

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: $$\frac{1}{W(1)}, \qquad \text{ where } W(z) \text{ is the LambertW function}$$

share|improve this answer
add comment

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: $$W(1)=W\left(\frac{1}{x}e^\frac{1}{x}\right)$$ $$W(1)=\frac{1}{x}$$ $$x=\frac{1}{W(1)}$$

share|improve this answer
add comment

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

share|improve this answer
add comment

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!

share|improve this answer
    
$L{}$ function? –  anorton Apr 8 '13 at 21:52
add comment

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

share|improve this answer
1  
(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. –  anorton 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. –  anorton Apr 8 '13 at 21:55
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.