Variable in both exponential and equation, trying to rewrite in terms of y

[jayce9936]

I started with the equation
x = y + e^[19.83 - 5417.75/(y+273.15)] - 5.56
My goal is to rewrite it in terms of y. Ive been playing around with it for quite a while seeing how I could rewrite it but it always ends up with y both as part of the equation as well as part of a natural logarithm or exponential. Currently ive got it down to:
(y+273.15)[ln(x-y+5.56)] = 19.83y - 1.1855
pretty clueless as to where to go from here. Dont have to do all the calculations for me just a push in the right direction would be appreciated.

 

[tkhunny]

There is no right direction. You have found a difficulty previously found by many. It is quite unsatisfying that it cannot be done.

All is not lost. There may still be numerical solutions. It doesn't have to be that complicated. [math]x = e^{x}[/math] is sufficient to have the same problem.

Now, when I say it cannot be done, I do not mean that we just haven't been sufficiently clever. I mean "cannot".

 

[jayce9936]

Lol good to know. Its the last step of a project so I guess I have to back and approach the whole thing from a different angle, but at least now I can stop wasting my time with something that has no solution.
Thank you for the insight

 

[LCKurtz]

Lol good to know. Its the last step of a project so I guess I have to back and approach the whole thing from a different angle, but at least now I can stop wasting my time with something that has no solution.
Thank you for the insight

Don't be confused by the difference between "no solution" and "no nice closed form expression" for the solution. Some expressions come up frequently enough that the "impossible" closed form expressions are given names and studied enough that the names and their values become as familiar as trig functions like [MATH]\sin(x)[/MATH]. To expand on tkhunny's example above, suppose you want to solve [MATH]4 = xe^{2x}[/MATH] for [MATH]x[/MATH]. Because problems like this arise so often, the Lambert [MATH]W[/MATH] function is defined as follows:
If [MATH]y = x e^x[/MATH], then the inverse function [MATH]x=W(y)[/MATH].
That is, the Lambert [MATH]W[/MATH] function is the inverse function of [MATH]y[/MATH]. What this means is that any equation that can be put in the form [MATH]y=xe^x[/MATH] can be solved for [MATH]x[/MATH] in terms of the Lambert W function. Here's how you would solve [MATH]4 = xe^{2x}[/MATH] using it. Multiply both sides by [MATH]2[/MATH] so [MATH]8 = 2xe^{2x}[/MATH]. This says [MATH]W(8)=2x[/MATH], so [MATH]x = \frac {W(8)} 2[/MATH]. The Lambert W is built into many software calculators and gives [MATH]x = .8029[/MATH]. I used Maple. I understand the online calculator DESMOS has it too, although I haven't checked it.

 

[jayce9936]

Don't be confused by the difference between "no solution" and "no nice closed form expression" for the solution. Some expressions come up frequently enough that the "impossible" closed form expressions are given names and studied enough that the names and their values become as familiar as trig functions like [MATH]\sin(x)[/MATH]. To expand on tkhunny's example above, suppose you want to solve [MATH]4 = xe^{2x}[/MATH] for [MATH]x[/MATH]. Because problems like this arise so often, the Lambert [MATH]W[/MATH] function is defined as follows:
If [MATH]y = x e^x[/MATH], then the inverse function [MATH]x=W(y)[/MATH].
That is, the Lambert [MATH]W[/MATH] function is the inverse function of [MATH]y[/MATH]. What this means is that any equation that can be put in the form [MATH]y=xe^x[/MATH] can be solved for [MATH]x[/MATH] in terms of the Lambert W function. Here's how you would solve [MATH]4 = xe^{2x}[/MATH] using it. Multiply both sides by [MATH]2[/MATH] so [MATH]8 = 2xe^{2x}[/MATH]. This says [MATH]W(8)=2x[/MATH], so [MATH]x = \frac {W(8)} 2[/MATH]. The Lambert W is built into many software calculators and gives [MATH]x = .8029[/MATH]. I used Maple. I understand the online calculator DESMOS has it too, although I haven't checked it.

yup thats what I ended up doing was graphing the inverse function. I only needed a specific value so I just read the point off the graph and put a screenshot in with my project to show how I arrived there.