The inverse of the equation $y=x^x$ is $\frac{\ln(x)}{W(\ln(x))}$. It is clear that the answer to $x^x=4$ is $2$, but the expression $\frac{\ln(4)}{W(\ln(4))}$ is not evidently equivalent with $2$. How can the expression be simplified to prove it is equal to 2?

To be more clear, $W(x)$ is the inverse of the function $y=xe^x$, so $x=W(y)$.

  • Do you accept your first sentence as true? Like do you question the first sentence. If not, we can say if $f(a)=b$ then $g(b)=a$. – randomgirl Dec 6 '17 at 2:02
  • I accept the first sentence as true, I know how to derive that from the original equation. – volcanrb Dec 6 '17 at 2:04
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    Did you want some way that doesn't use the fact that the functions are inverses to show that $\frac{\ln(4)}{W(\ln(4))}=2$? – randomgirl Dec 6 '17 at 2:08
  • Provide your definition of $W$. If this definition is the inverse of a particular equation involving $xe^x$, then use that definition... should amount directly to showing that $2^2=4$... – Simply Beautiful Art Dec 6 '17 at 2:09
  • What I'm looking for is basically, if you had no idea that 2 was the answer to the equation $x^x=4$, how would you find that the answer is 2? And I thought using the inverse may be a way to go about that but maybe it is not the best way. – volcanrb Dec 6 '17 at 2:22
up vote 2 down vote accepted

In order to simplify your expression you require the following simplification rule for the Lambert W function $$\ln x = \begin{cases} \text{W}_0 (x \ln x), \quad x \geqslant \dfrac{1}{e},\\[2ex] \text{W}_{-1} (x \ln x), \quad 0 < x \leqslant \dfrac{1}{e}. \end{cases}$$ Note here $\text{W}_0 (x)$ denotes the principal branch for the Lambert W function while $\text{W}_{-1} (x)$ corresponds to the secondary real branch.

Now, for $\text{W}_0 (\ln 4)$, on applying the above simplication rule we have $$\text{W}_0 (\ln (4)) = \text{W}_0 (2 \ln 2) = \ln 2.$$ Thus $$\frac{\ln (4)}{\text{W}_0 (\ln (4))} = \frac{2 \ln 2}{\ln 2} = 2,$$ as expected.

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