Mathematics Weblog

The LambertW Function

Friday 9 April 2004 at 2:35 pm | In Articles | 1 Comment

Many equations cannot be solved exactly without using special functions. For example, to solve 2^x=3 requires the use of the $\ln$ function (or similar). This function is sometimes defined in terms of an integral from which their properties can be deduced. Thus $\ln$ is defined by $\displaystyle \ln x=\int ^x _1 \dfrac{dt}{t}\text{ for } x>0$ and it is then clear that, for example, $\ln 1=0$

There are many equations that can only be solved in terms of newly-defined functions. One such function that isn’t all that well known is the LambertW function where $w\left(x\right)$ is defined as a solution (for ) of we^w=x . This allows you to solve equations like 2^x=x^8 which was asked about on the S.O.S. Mathematics CyberBoard

To solve 2^x=x^8 let $w=-\ln x$ so that $x=e^{-w}$ . Then

$2^x=x^8 \Longrightarrow 2^{e^{-w}}=e^{-8w} \Longrightarrow e^{-w}\ln 2=-8w \Longrightarrow -\frac{1}{8}\ln 2=we^w$

Thus $w=w(\frac{1}{8}\ln 2)$ and so $x=e^{-w}=-8\dfrac{w(\frac{1}{8}\ln 2)}{\ln 2}$ which is our answer.

Using tables or software this gives 1.100.

But hang on, is that the only solution? No, because 2^x<x ^8 for small values of and 2^x grows much faster than x^8 so 2^x>x^8 for large values of . Since both $x \mapsto 2^x$ and $x \mapsto x^8$ are continuous on $\mathbb{R}$ there is another value of for which 2^x=x^8 . A quick fiddle with a calculator gives $x\approx43.5$ .