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Can anyone help me solve e^x-ex-e=0?

9 Answers

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Oof, algebraically? Not an easy route. How about you solve by graphing? The values of x that make a function equal to zero are its x-intercepts.
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If you don't know about Lambert W functions, you probably aren't going to find an answer that satisfies you.

There isn't an analytical solution (without using Lambert W functions).

You can solve it numerically if you're okay with that
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do you have a picture of the original question?
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Likely missing some rigour that hopefully someone else can fill in.

Let g(x) = e\^x - ex - e

We note that lim x--> -infinity g(x) = infinity, lim x --> infinity g(x) = infinity.

We can also show that g'(x) = e\^x - e is just e\^x moved down by e units, so it has exactly 1 zero.

With only value satisfying g'(x) = 0 on continuous, differentiable g, g(x) is U shaped.

We can show that there exist some continuous domain on g, such that g(x) < 0. For example, g(0) = -e < 0.

By intermediate value theorem we obtain g(x) has exactly two zeros.
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The Lambert W function should come in handy
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Have you had Calculus I? You could use Newton's method here.
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Please let us know what the teacher says. Imo I think it's just a mean question to make you think, where they didn't expect any concrete answer from you.
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You can divide everything by e and it will be easier to work with.

Lol...getting downvoted when it's perfectly reasonable as a simplification.
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First of we assume there is a function: W( x e \^ x ) = x.  
Then the rest is easy.   

e \^ x - e x - x = y   
*( y instead of 0 for generalizing )*   
 
e \^ ( x - 1 ) - x - 1 = y ÷ e   
*( we now recognize that exponent is similar to the other terms, so we try to make it fit in the W  format )*

e \^ ( e \^ ( x - 1 ) ) e \^ ( - x - 1 ) = e \^ ( y e \^ ( - 1 ) )   
*( since the latter factor needs to be the same as the first exponent, we add 2 and multiply the exponent with -1 and multiply everything with -1 )*

e \^ ( - e \^ ( x - 1 ) ) ( - e \^ ( x - 1 ) ) = - e \^ ( - ( 2 + y e \^ ( - 1 ) )   
*( now it fits with W )*

( - e \^ ( x - 1 ) ) = W( - e \^ ( - ( 2 + y e \^ ( - 1 ) ) )   
*( multiplying with -1, taking ln and adding +1)*

x = ln( - W( - e \^ ( - ( 2 + y e \^ ( - 1 ) ) ) ) ) + 1   
*( substituting y with 0 )*

ln( - W( - e \^ ( - ( 2 + 0 e \^ ( - 1 ) ) ) ) ) + 1

ln( - W( - e \^ ( - 2 ) ) ) + 1 or ln( -W( -e^-2 ) )

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