@Ian Fowler Any solution will be ok even if it uses either lambert W function or some non - elementary function. Until the solution is computable it is okay for me.

I hate to be the fly in the ointment again, but here's throwing caution to the wind.

In line 15 you have some of the form : e^2y = B.

In log form this become: 2y = ln(B),

not 2y = B which what you have in line 17

You can even take the route of e^2y = e^ln(B) , which is unnecessary, but it still leads to 2y = ln(B). So line 17 is missing the ln function.

The bottom line is you have 2y = ln( a function of y) which does not have an elementary solution. Maybe there is some kind of Lambert function solution but that's the best you have.

Wasi: You might try using the infinite Maclauren series for arc_tanh(x) and use a computer to grind out some desired level of accuracy and then go from there or the Lambert function possibility. I think those 2 are your only shot.

Faizan: I don't mean to flog the point but you missed the 2nd part of my response.

In line 15 you have: e^2y = B where B = [sqrt(m) +t] / [sqrt(m) - t]

Then on line 17 you have : 2y = B

This is incorrect as line 15 implies that: 2y = ln(B)

So everything that follows doesn't work.

Wasi: I have discovered another problem:

line 15 is O.K.

I assume p is a density, A is a cross-sectional? area and x is a length?

At any rate, assuming m,p,A are positive line 15 does not have a real solution for positive x as [sqrt(m) +sqrt(m+pAx] / [sqrt(m) - sqrt(m+pAx)] will be negative since

Sorry Faizan. In your solution, 2y = ln[sqrt(m) +t] / [sqrt(m) + t] leads to 2y = ln(1) = 0

So y = 0 leads to 2t - 1 = 0 or t = 1/2

That leads to x = [1-4m]/4pA.

But it's worse than that because I think you have made a sign error in line 13:

e^2y = ln { [sqrt(m) + t] / [sqrt(m)

-t)] } - the green "-"You will not be able to isolate t. Too bad about the minus sign. Otherwise it would have worked out.

Wasi: tanh[f(y)] = g(y) cannot be solved for y unless f(y) is a constant. Maybe there is some kind of convoluted Lambert function solution.

That's my take, unless I have missed something obvious.

@Ian Fowler Any solution will be ok even if it uses either lambert W function or some non - elementary function. Until the solution is computable it is okay for me.

@Ian Fowler

Thanks for pointing out the typo error The updated solution is here.

I hate to be the fly in the ointment again, but here's throwing caution to the wind.

In line 15 you have some of the form :

e^2y = B.In log form this become:

2y = ln(B),not 2y = B which what you have in line 17

You can even take the route of e^2y = e^ln(B) , which is unnecessary, but it still leads to 2y = ln(B). So line 17 is

missing the ln function.The bottom line is you have 2y = ln( a function of y) which does not have an elementary solution. Maybe there is some kind of Lambert function solution but that's the best you have.

Wasi: You might try using the infinite Maclauren series for arc_tanh(x) and use a computer to grind out some desired level of accuracy and then go from there or the Lambert function possibility. I think those 2 are your only shot.

Faizan: I don't mean to flog the point but you missed the 2nd part of my response.In line 15 you have:

where B = [sqrt(m) +t] / [sqrt(m) - t]e^2y = BThen on line 17 you have :

2y = BThis is incorrectas line 15 implies that:2y = ln(B)So everything that follows doesn't work.

Wasi:I have discovered another problem:line 15 is O.K.

I assume p is a density, A is a cross-sectional? area and x is a length?

At any rate, assuming m,p,A are positive line 15[sqrt(m) +sqrt(m+pAx] / [sqrt(m) - sqrt(m+pAx)] will be negative since

does not have a real solution for positive xassqrt(m+pAx) > sqrt(m).