## nshiba 4 years ago e^x=-x+1, how do you solve this? @MIT 18.01 Single …

1. JamesJ

In this case x = 0 is the solution.

2. JamesJ

Normally with an expression e^(ax) = polynomial in x, numerical methods are needed. But with yours we can see a solution by inspection.

3. arijit.mech

friend it is easy:just draw the graphs of y=e^x & y=-x+1....>the intersection point is the solution....this is a transcendental equation |dw:1320213287494:dw|

4. rosswante
5. arijit.mech

Thanks....i jst give u a hint

6. verma.milan

mathematically take natural log on both sides e u will get point x where the solution exists

7. verma.milan

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8. alireza_masoomi

dif ratio x then Ln that result x=e^-1

9. alireza_masoomi

hi! e^x=-x+1when dif it then e^x=-1 after Ln xLne=Ln-1 that result x=e^-1 ~ .36

10. alireza_masoomi

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11. JamesJ

The equation \[ e^x = -x + 1 \] has one and only solution, \( x = 1. \ \ \ x = e^{-1} \) is not a solution, something that is easily verifiable by substitution. (The uniqueness of the solution \( x = 1 \) can be proven by examining the function \( f(x) = e^x + x - 1. \ \ \ x = 1 \) is clearly a zero of f, but as \( f'(x) = e^x + 1 > 0 \) for all \(x\), it follows that it is also the only zero.)