How to solve for a variable in an exponential equation

I want to solve this function for rs.. I used this code

syms ra rb rs Vr

Eq1 = Vr == exp((-pi*ra)/rs)+exp((-pi*rb)/rs);

t_sol = solve(Eq1, rs)

but I am getting an empty sym can anyone help me it's urgent

[Walter Roberson]

http://www.mathworks.com/matlabcentral/answers/29922-why-your-question-is-not-urgent-or-an-emergency

Answer 1

[Walter Roberson]

There is no closed form solution for that.

In Maple notation,

rs := -Pi * rb / RootOf(-exp(_Z)+Vr-exp(ra*_Z/rb))

Here, RootOf(-exp(_Z)+Vr-exp(ra*_Z/rb)) stands in for the set of values, _Z, such that the expression inside RootOf() becomes zero -- the roots of the equation. However, there is no closed form expression to derive that set.

[Alice Faisal]

Z stands for what Actually, I have values for the other parameters and I need to get rs. But when computing it on MATLAB I dont get a correct answer This is the equation

 exp((-pi*ra)/rs)+exp((-pi*rb)/rs)=1

and ra=90.82*10^6 rb=90.23*10^6 so only rs in unknown

[Walter Roberson]

In Maple notation RootOf(expression involving _Z) stands in for all of the values for the variable _Z such that the expression becomes zero. It is, in other words, a place holder for the roots of an equation.

[Walter Roberson]

4.1029039177055862347725735102730*10^8 approximately. Possibly other solutions as well, probably complex solutions.

[Alice Faisal]

Yes this is the correct answer, how did you get it using matlab? Because I need to try other vales for ra which is 2.30 and rb =6.89 Thank you so much in advance

[Walter Roberson]

Assuming you mean 2.30E6 and 6.89E6 then 1.8887251678153105468716736113549*10^7

I used a different package, but I did the equivalent of using vpasolve() with an initial value of rs = 500000000