Lambert W Function – Impossible math equation?

Many problems which looked impossible have been solved using the Lambert W Function. Here we solve on such problems where pre closed-form solutions are impossible. Here using the Lambert W Function we can get the solution with the help of precise algebraic corrections for the dominant terms. The problem is given below.

Problem – Lambert W function ?

Find a+b=? if 7^a+b^7=a^b

At first glance it looks as if it is an impossible math equation. Here the article is about solving the above impossible math equation step by step by using Lambert W function.

Solution

3D image of solution

As already discussed it is a very complex problem requiring lot of in depth knowledge of Lambert W Function and other parameters. Let us have a look at the 3D image of the solution and then solve the problem.

Step by Step Solution

The given problem involves an equation with two variables a and b and we have to find out a+b=?. Here the problem to be solved is of one equation with two variables. Hence it is not an ordinary problem.

The problem is to find a+b if 7^a+b^7=a^b .

Here we are required to find out the dominant terms in the given equation. Let us assume that a>0 and b>0.

From the given equation, let us say, f(b)=a^b-b^7-7^a=0

Using the principles of dominant terms, the reduced equation will be a^b=b^7

Hence, a^(b/7)=b and blna=7lnb

From above,

$\begin{array}{c}\frac{\ln a}{7}=\frac{\ln b}{b}\\ \frac{1}{b}=e^{-\ln b}\\ -\frac{\ln a}{7}=-\ln b.e^{-\ln b}\end{array}$

The above is in the general form of Lambert W Function, W(ue^u)=u, we get,

$W(-\frac{\ln a}{7})=-\ln b$ , comparing it with $e^{-W(x)}=\frac{W(x)}{x}$ for $b_0$ .

$b_0=\frac{W(-\frac{\ln a}{7})}{-\frac{\ln a}{7}}$

Application of calculus – first order derivative

From the dominant terms assumption, we get $a^{b_0}=b_0{}^7$, hence,

$f(b_0)=a^{b_0}-b_0{}^7-7^a=b_0{}^7-b_0{}^7-7^a=-7^a$

$f^{\prime }(b)=a^b\ln a-7b^6$

$f^{\prime }(b_0)=b_0{}^7\ln a-7b_0{}^6$

$f^{\prime }(b_0)=b_0{}^7\ln a-7b_0{}^6=b_0{}^6(-7W-7)=-7b_0{}^6(W+1)$

Fro above,

$b\approx b_0-\frac{7^{a-1}}{7b_0{}^6(W+1)}$

In the above equation, the value of a is finalized on the basis of the inequalities applied as per algebraic conditions of the given problem.

For 1<a<13.13, the $W_{-1}\mathrm{branch}$ the large root and $W_0\mathrm{branch}$yields the small root.

Hence when a=2, b=36.263

The above is one of the solutions. We can get more solutions for the possible values of a as pr the given problem.

It will be interesting to check that at a=20, b= 13. Hence there are no unique solutions abut not for any arbitrary values of a.

At $a=e^{(7/e)},b_0\ln a-7=0$ .

Hence the chosen value of a cannot be arbitrary. It should be defined and as per the given problem conditions.

Thus the above problem solution is very complex and requires thorough knowledge of Lambert W Function and its branches.

The video solutions for reference are available from you tube channel maths&science by iitian at :

https://www.youtube.com/@mathsciencebyiitian

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Best of luck to everyone.