HELP! solving simple system of equations

kochibacha said:
1-ke^-t=3

2t-ke^(-3t)=4

could you help me do this i cant really solve this
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The first equation can be written \(\displaystyle -ke^{-t}= 2\) so that \(\displaystyle k= -2e^t\). Putting that into the second equation gives \(\displaystyle 2t+2(-2e^t)(e^{-3t})= 2t- 4e^{-2t}= 2\) or \(\displaystyle t- 2e^{-2t}= 1\). That cannot be solved in terms of "elementary" functions but can be put into a form so that the "Lambert W function" (the inverse function to \(\displaystyle f(x)= xe^x\)) can be used.
 
Of course, there is an exact solution. There just isn't any easy way to write it (if you don't consider the "Lambert W function" easy). For example, an exact solution to \(\displaystyle x^2= 2\) is \(\displaystyle \sqrt{2}\) but that's only because that is the way "square root" is defined. If you did not have a function defined to be the solution to that equation, you could not write that exact solution.

Here, as I said before and as corrected by Denis, from \(\displaystyle -ke^{-t}= 2\), we get \(\displaystyle k=-e^t\). Putting that into the second equation, \(\displaystyle t+ 2e^{-2t}= 2\) so that \(\displaystyle 2e^{-t}= 2- t\). Let u= 2- t so that t= 2- u and the equation become \(\displaystyle 2e^{u- 2}= 2e^2e^u= u\) and the \(\displaystyle ue^{-u}= 2e^2\). Finally, let v= -u so that we have \(\displaystyle -ve^v= 2e^2\) or \(\displaystyle ve^v= -2e^2\). The solution to that equation is \(\displaystyle v= W(-2e^2)\) where W is the lambert W function. Then \(\displaystyle u= -W(-2e^2)\) and \(\displaystyle t= W(-2e^2)+ 2\). That is an "exact solution" in the same sense that \(\displaystyle \sqrt{2}\) is an exact solution to \(\displaystyle x^2= 2\) and \(\displaystyle ln(3)\) is the exact solution to \(\displaystyle e^x= 3\). None of those is a rational number so we cannot write it as a terminating or repeating decimal but that is NOT what is meant by "exact solution".
 
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HallsofIvy said:
Of course, there is an exact solution. There just isn't any easy way to write it (if you don't consider the "Lambert W function" easy). For example, an exact solution to \(\displaystyle x^2= 2\) is \(\displaystyle \sqrt{2}\) but that's only because that is the way "square root" is defined. If you did not have a function defined to be the solution to that equation, you could not write that exact solution.

Here, as I said before and as corrected by Denis, from \(\displaystyle -ke^{-t}= 2\), we get \(\displaystyle k=-e^t\). Putting that into the second equation, \(\displaystyle t+ 2e^{-2t}= 2\) so that \(\displaystyle 2e^{-t}= 2- t\). Let u= 2- t so that t= 2- u and the equation become \(\displaystyle 2e^{u- 2}= 2e^2e^u= u\) and the \(\displaystyle ue^{-u}= 2e^2\). Finally, let v= -u so that we have \(\displaystyle -ve^v= 2e^2\) or \(\displaystyle ve^v= -2e^2\). The solution to that equation is \(\displaystyle v= W(-2e^2)\) where W is the lambert W function. Then \(\displaystyle u= -W(-2e^2)\) and \(\displaystyle t= W(-2e^2)+ 2\). That is an "exact solution" in the same sense that \(\displaystyle \sqrt{2}\) is an exact solution to \(\displaystyle x^2= 2\) and \(\displaystyle ln(3)\) is the exact solution to \(\displaystyle e^x= 3\). None of those is a rational number so we cannot write it as a terminating or repeating decimal but that is NOT what is meant by "exact solution".
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So how can i translate the v= W(-2e^2) into the value that we can plug into the equation to calculate the prediction or result from the modeling I have made (such as v=1.25143...
I thinks they are always an irrational number because of the exponential function)
 
kochibacha said:
anyone?
Click to expand...

If you're now asking for decimal approximations (for k and t), then you could use software (like Denis did).

Evaluating the LambertW function, using a program called MapleV, yields the following approximations (rounded).

k = -14.49960147

t = 1.980973984

OR

k = -1.278394533

t = -0.447542161

Cheers :cool:
 
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