The Number e and natural log

e^(x+1) + e^x = 2
e^x + e^1 + e^x = 2
e^x(e^1 + 1) = 2
e^x(2.7183 + 1) = 2
e^x(3.7183) = 2
e^x = 2/3.7183
e^x = 0.5379
x = ln 0.5379
x = -0.6201

Check: Let x = -0.6201
e^(x+1) + e^x = 2
e^(-0.6201+1) + e^(-0.6201) = 2
e^0.3799 + e^(-0.6201) = 2
1.4621 + 0.5379 = 2
2 = 2

2010-08-31 02:09:58 補充：
They want the answer in exact value.

2010-08-31 02:16:46 補充：
sorry that line 2 is wrong
e^x + e^1 + e^x = 2

The correct one is (e^x)(e^1) + e^x = 2

because e^(x+1) = e^x e^1

The rest are OK.

003: have you studied amaths yet?
It's true for the existence of natural number "e".

2010-08-31 13:48:44 補充：
Please refer to
http://en.wikipedia.org/wiki/E_(mathematical_constant)

in terms of e

e^(x+1)+e^x=2
e*e^x+e^x=2
(e+1)e^x=2
e^x= 2/(e+1)
x=ln[ 2/(e+1) ] or ln(2)-ln(e+1)

邊有exact value??
又e 又x.....

2010-08-31 13:14:51 補充：
002
e^1 = 2.7183
邊個講架?

你老吹??