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nshiba
e^x=-x+1, how do you solve this? @MIT 18.01 Single …
In this case x = 0 is the solution.
Normally with an expression e^(ax) = polynomial in x, numerical methods are needed. But with yours we can see a solution by inspection.
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|
You may see http://www.wolframalpha.com/input/?i=solve%7Be%5Ex%3D-x%2B1%7D
Thanks....i jst give u a hint
mathematically take natural log on both sides e u will get point x where the solution exists
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dif ratio x then Ln that result x=e^-1
hi! e^x=-x+1when dif it then e^x=-1 after Ln xLne=Ln-1 that result x=e^-1 ~ .36
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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.)