Why is that restriction needed? Why are not we allowed to take positive c? I can see that c=0 is impossible as x=ln(1+e^x) means e^x=1+e^x or ′0=1′.

Amira Serrano

Amira Serrano

Answered question

2022-10-19

Why x ln ( 1 + e x ) = c has a solution for every c < 0 and not otherwise?
Why is that restriction needed? Why are not we allowed to take positive c? I can see that c = 0 is impossible as x = ln ( 1 + e x ) means e x = 1 + e x or 0 = 1 .

Answer & Explanation

Teagan Zamora

Teagan Zamora

Beginner2022-10-20Added 18 answers

x = ln ( 1 + e x ) + c
e x = ( 1 + e x ) e c
e x = e c 1 e c = 1 1 e c 1
Now, c < 0 0 < e c < 1 0 < 1 e c < 1 1 1 e c > 1 Since the equation e x = k is solvable iff k > 0, we are done.
Jaylyn Horne

Jaylyn Horne

Beginner2022-10-21Added 5 answers

Because as x , we have f ( x ) = x l n ( 1 + e x ) 0 , meaning that the x axis is a horizontal asymptote. By differentiating and observing the derivative, you can see that the function is monotonic increasing, meaning that the 'highest' that the curve ever gets is infinitely close to the x axis.
So if you draw y = c, where c > 0, it will NOT intersect the function, which is why x l n ( e x + 1 ) = c has no solutions. However, when c < 0, there will be solutions because y = c would then be below the x axis, where it will intersect the function.

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