Solving ##a^b## and ##b^a## via the Lambert Function

[jedishrfu]

The Lambert function comes up in a lot of math videos. Here's one such video:

 

[PeroK]

The Lambert function comes up in a lot of math videos. Here's one such video:

In school, they never taught "proof by inspection" either!

 

[anuttarasammyak]

The Lambert function comes up in a lot of math videos. Here's one such video:

Since $4^x + x$ is monotonically increasing, I would recommend that students make a rough estimation; they would then find that $4^x + x=4^4+4$

 

[jedishrfu]

But the point of the video is to teach us how to use the lambert function to solve a problem not how to intelligently guess the answer.

In the case of the original expression here, students might guess similarly setting a=b and yet miss 2^4-4^2 solution.

 

[PeroK]

But the point of the video is to teach us how to use the lambert function to solve a problem not how to intelligently guess the answer.

After all the complications with applying the W-function, the solution still hinges on noticing that $4^4 = 256$. You might as well notice that right up front.

 

[PeroK]

But the point of the video is to teach us how to use the lambert function to solve a problem not how to intelligently guess the answer.

In the case of the original expression here, students might guess similarly setting a=b and yet miss 2^4-4^2 solution.

Here's an analysis. We start with a more general equation:$$a^x + x = b$$$$(b - x)a^{-x} = 1$$$$(b-x)a^{b-x} = a^b$$We want to use the W-function on the left-hand side, so we use that $y = e^{\ln y}$ and multiply both sides by $\ln a$:$$(b-x)(\ln a)e^{(b-x)\ln a} = a^b \ln a$$Now, by the definition of the W-function:$$(b-x)\ln a = W(a^b \ln a)$$And, we have the general solution:$$x = b - \frac{W(a^b \ln a)}{\ln a}$$To get rid of the W-function, we need to factorise $a^b = a^aa^{b-a}$ and use the trick that $y = e^{\ln y}$ again:$$x = b - \frac{W((a^a \ln a)e^{(b-a)\ln a})}{\ln a}$$So, in the special case that $a^a = (b - a)$, we have:$$x = b - (b-a) = a$$
Let's go back to the general solution:$$x = b - \frac{W(a^b \ln a)}{\ln a}$$We try to get rid of the W-function by factorising $a^b = a^{\alpha}a^{b-\alpha}$ to give:$$x = b - \frac{W((a^\alpha \ln a)e^{(b - \alpha)\ln a})}{\ln a}$$Then, in order to simplify we need to solve:$$a^\alpha = b - \alpha$$Which is precisely our original equation!

So, all the shenanigans with the W-function was going round in one big circle to get us back to our original equation within the W-function argument. The only way to simplify is to spot/guess the solution to the original equation!

That said, it's still worthwhile to get a general solution in terms of the W-function. But, it doesn't help to get a specific solution.

I had a gut feeling there was something not right about that video.