Solving A Homemade Eulerian Exponential Equation

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[meve]

Rearranging the original equation, x = 1- e^(1-e^x). That is x = f(f(x)) with f(x) being 1-e^x. f(x) is a decreasing function so f(f(x)) is an increasing function but f^(-1)(x)=f(x) only when x=0 (obtained by guess and check) so x=0 is the only solution.

[abdoshaat]

i think it's better If there's no algaberic solutions you can inform us in the beginning instead of this analytic guess and check ‏‪

[heman___]

I dont understand why everyone is making this easy question too complex.

Solution :-
x+e^(1-e^x)=1....(i)
==> x+e/e^(e^x)=1
[using a^(m-n)=(a^m)/(a^n)
==> (x.e^(e^x) + e) / e^(e^x) = 1
[taking l c m]
==> x.e^(e^x) + e = e^(e^x)
[multiplying lhs and rhs by e^(e^x) or we can say sending e^(e^x) to rhs]
==> e^(e^x){x+e^((e^x)-1)=e^(e^x)
[taking e^(e^x) common and then cancelling out it in lhs and rhs, we get :- ]
x + e^((e^x)-1)=1....(ii)
Comparing eq (i) and (ii), we get :
e^(1-(e^x))=e^(e^x-1)
Since bases are same, therfore :
1-e^x=e^x-1
==> 2=2.e^x
==> e^x = 1
==> x=0
as simple as it ; )

[allanmarder]

By guess and check x=0 is a solution. The derivative of f(x) = x + e^(1-e^x) -1 is 1+ (e^(1-e^x)(*(-e^x) = 1-e^(1+x - e^x). The taylor series for
e^x is 1+ x+ (x^2)/2! + (x^3)/3! +.... So that 1+x -e^x = -(x^2)/2! -(x^3)/3! - ..and is <0 for x>0. The series can also be written as
1+x -e^x = -((x^2)/2! )*(1-(x/3)) -((x^4)/4! )*(1-(x/5)) and is less than 0 for x<0. Thus (1+x - e^x) is <0 for all x.
and 1- e^(1+x - e^x) > 0. for all x and the derivative of f(x) is >=0. So f(x) is increasing and x=0 is the only solution.

[mathsandsciencechannel]

THIS IS AN INTERESTING EULER EXPONENTIAL EQUATION. THE AI SAID IT CAN ONLY BE SOLVED THIS WAY:
WELL DONE SYBERMATH
The equation you provided is:

x + e^(1 - e^x) = 1

Unfortunately, this equation cannot be solved analytically using elementary functions. However, we can attempt to find an approximate solution using numerical methods or graphical analysis.

One approach is to plot the graph of the equation and observe where it intersects the line y = 1. We can use a graphing calculator or software to visualize the equation and estimate the solutions.

[NXT_LVL_DVL]

Can you add the Lambert W version of the problem ?

[peterromero]

This one is very hard to follow! Thanks for working through it.

[Luffy_wastaken]

Can you make one on why cos(sinx)>sin(cosx) always?

[gdtargetvn]

e^(1-e^x)=1-x
ln both sides: 1-e^x=ln(1-x)
Let f(x)=1-e^x
e^x=1-f(x)
x=ln(1-f(x)), this is f^(-1)(x)
Basically f(x)=f^(-1)(x)
=> f(x)=x
Solve 1-e^x=x
LHS is decreasing, RHS is increasing, they intercept at only 1 point, (0, 0)
Therefore, x=0

[Ghaith]

ohh so thats whats subbstituion is for hmmm... intresting. awesome

[moeberry]

This is just an equation of the type f=f inverse.

[nicolascamargo]

Se tiene x=1-e^(1-e^x) que tal si f(x)=1-e^x entonces x=f(1-e^x) por lo tanto f(1-e^x)=f(x) de donde 1-e^x=x lo cual es equivalente a e^x=1-x lo cual como e^x es una función estrictamente creciente y 1-x es estrictamente decreciente solo hay un punto de corte que probando x=0 es donde se da por lo tanto es la única solución.

[sngmn]

Increasing... decreasing..., guess and check, x = 0. Done)

Edited: not so fast guys, im not right actually.

[angelishify]

Enačbo odvajamo: 1 - e^x * e^(1 - e^x) = 0 ; e^(x + 1 - e^x) = 1 ter e^x = x + 1. Premica y = x + 1 se v točki x = 0 dotika exponentne funkcije y = e^x .

[sngmn]

Wolfram alpha understand f(x) = e^(1-e^x) but not f(x) = 1 - e^(1-e^x) LOL.

[pzelact]

when a*x*lnx and a^(x*lnx) has 1 common point

[someperson]

Let f(x) =: 1 + x - e^x, for real x. Then, f'(x) = 1 - e^x. Thus, f'(x) = 0 iff x = 0. Since f'(-1) > 0 and f'(1) < 0, it follows that f(x) <= f(0) = 0 and (f(x) < 0, for x <> 0). Le t
g(x) =: x + e^(1 - e^x). Then, g'(x) = 1 - e^f(x). So, g'(x) <= 0 and (g'(x) = 0 iff x = 0). Hence,
g is strictly decreasing. Therefore, g(x) = 1 has at most one solution. Also, g(0) = 1.

[angelishify]

x = 1 - e^(1 - e^x); za x vstavimo desno stran in dobimo: x = 1 - e^(1 - e^ (1 - e^(1 - e^x))). To lahko ponavljamo v neskončno: x = 1 - e^(1 - e^ (1 - e^(1 - e^(...)))). Če stvar konvergira, to pomeni, da je tudi (1 - e^ (1 - e^(1 - e^(...)))) = x in zato x = 1 - e^x.