\({e}^{-10/3x}+{e}^{-7/3x}+{e}^{-x}=0.15\) , how to solve this ????? find x.
Are you sure you have copied it correctly?
Do you mean
\({e}^{(-10/3)x}+{e}^{(-7/3)x}+{e}^{-x}=0.15\qquad(1)\\~\\ or\\~\\ {e}^{-10/(3x)}+{e}^{-7/(3x)}+{e}^{-x}=0.15\qquad (2)\)
the first equation is right, the answer for x is 1.97568, but how can u get this?
As I said in the answer that the Math Package that I have used "iteration method" to get that answer. "Iteration method" is a fancy phrase for "trial and error method"!!. The computer "guesses" and then tries to see if the equation balances, or how close it is to the real answer and repeats the process through iteration and interpolation, until it finds the "right" answer that balances the equation.
It does this very rapidly, since it can do hundreds or even thousands of these iterations in one second.
Translation: stick your d i c k in the whatcha macallit and see what happens.
This is Mr. BB. Gingerale trolls this f u c k face because he is so stu.pid and dum.b he is actually contagious.
A long time ago, Mr. BB stuck his d ic k ina whatcha macallit that was actually a dirtdevil vacuum. These have the fan blade in the front. Now Mr. BB has to sit when he takes a p i ss
I read your question as: ({e})^(-10/3 x) + ({e})^(-7/3 x) + ({e})^({-x}) = 0.15 , as Melody suggested in Equation (1).
"Mathematica 11 Home Edition" gives this answer for x, by iteration only. It does not give any step by step solution.
x = 1.97568 . This value of x does balance the equation as written above.
If it is Equation(2) as suggested by Melody, then there is no "real" solution for x. There is, however, a complex solution, and "Mathematica 11" gives this complex value for x:
x ≈ 0.651484 - 0.910470 i . But no explanation! Sorry about that.