Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing

This Question is Closed

JamesJ Group TitleBest ResponseYou've already chosen the best response.0
In this case x = 0 is the solution.
 3 years ago

JamesJ Group TitleBest 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.
 3 years ago

arijit.mech Group TitleBest 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
 3 years ago

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

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

verma.milan Group TitleBest 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
 3 years ago

verma.milan Group TitleBest ResponseYou've already chosen the best response.0
dw:1321869467155:dw
 3 years ago

alireza_masoomi Group TitleBest ResponseYou've already chosen the best response.0
dif ratio x then Ln that result x=e^1
 3 years ago

alireza_masoomi Group TitleBest 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
 3 years ago

alireza_masoomi Group TitleBest ResponseYou've already chosen the best response.0
dw:1322221990014:dwdw:1322222039311:dw
 3 years ago

JamesJ Group TitleBest 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.)
 3 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.