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JamesJBest ResponseYou've already chosen the best response.0
In this case x = 0 is the solution.
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
Normally with an expression e^(ax) = polynomial in x, numerical methods are needed. But with yours we can see a solution by inspection.
 2 years ago

arijit.mechBest ResponseYou've already chosen the best response.2
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
 2 years ago

rosswanteBest ResponseYou've already chosen the best response.0
You may see http://www.wolframalpha.com/input/?i=solve%7Be%5Ex%3Dx%2B1%7D
 2 years ago

arijit.mechBest ResponseYou've already chosen the best response.2
Thanks....i jst give u a hint
 2 years ago

verma.milanBest ResponseYou've already chosen the best response.0
mathematically take natural log on both sides e u will get point x where the solution exists
 2 years ago

verma.milanBest ResponseYou've already chosen the best response.0
dw:1321869467155:dw
 2 years ago

alireza_masoomiBest ResponseYou've already chosen the best response.0
dif ratio x then Ln that result x=e^1
 2 years ago

alireza_masoomiBest ResponseYou've already chosen the best response.0
hi! e^x=x+1when dif it then e^x=1 after Ln xLne=Ln1 that result x=e^1 ~ .36
 2 years ago

alireza_masoomiBest ResponseYou've already chosen the best response.0
dw:1322221990014:dwdw:1322222039311:dw
 2 years ago

JamesJBest ResponseYou've already chosen the best response.0
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.)
 2 years ago
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