How to find the exact answer of an exponential equation?

syms t 
solve(exp(-0.04*t)+exp(-0.12*t)==1,t)

I got the answer for this but I don't understand what it meant? I need a time value for this. Can someone explain me what this happened and how to get the exact value of t?

25*log(root(z^3 - z^2 - 1, z, 1))

Thank you so much!!

Answer 1

[Rik]

You can use double to extract a numerical value, but if you're looking for a time, something is going wrong here. The numerical approach shows there actually is a solution.

syms t 
solve((exp(-0.04*t)+exp(-0.12*t))==1,t),  double(ans)

ans = 

ans = -4.7781 - 46.4120i

f=@(t) exp(-0.04*t)+exp(-0.12*t);
sol_t=fminsearch(@(t) abs(f(t)-1),10), f(sol_t)

sol_t = 9.5562

ans = 1.0000

Answer 2

[John D'Errico]

syms t real
Eqn = exp(-0.04*t)+exp(-0.12*t) - 1;

Write it like that. I did so because now we can plot it. See that I specified t to be a real variable, since you are not interested in complex valued solutions.

Now, we can plot it. Does a solution exist? Are there multiple solutions? ALWAYS PLOT EVERYTHING. Look at what you see. Think about what you see. Then plot it in a different way if necessary. Where this relationship crosses zero, this is your solution.

fplot(Eqn,[-5,25])
yline(0)
grid on

So a solution does exist, and I would bet there is only one real solution. The one you care about is a little less than 10.

tsol = solve(Eqn,t)

tsol = 

And that may not seem terribly useful, but it is. The result is the third root of a cubic polynomial, You need to push MATLAB to resolve the solution. VPA will do that here.

vpa(tsol)

ans = 

9.5561271460008910332339624796214

You could also have gone directly to a numerical solution using vpasolve.

vpasolve(Eqn)

ans = 

9.5561271460008910332339624796214

In fact, that polynomial had three roots, but two will yield complex values. If you wish an algebraic solution for the problem, you could do this:

solve(Eqn,t,'maxdegree',3)

ans =