# solve

posted by .

e^c-2c = e-2
what is c?

ummm, 1?

e^1 = e
2*1 = 2

How do I get this?

c = 1 is one solution. To find the other, you can proceed as follows.

e^c-2c - e + 2 = 0

put f(c) = e^c-2c - e + 2

Newton's Method (approximate functon by the tangent) gives succesive better and better approximations.

c_{n+1} = c_{n} - f(c_{n})/f'(c_{n})

= c_{n} -
[e^c_{n}-2c_{n} - e + 2]/[e^c_{n} - 2]

Take c_{1} = 0

then:

c_{2}=0.2817

c_{3}=0.3465

c_{4}= 0.351326

c_{5}= 0.351354757134

c_{6}= 0.351354758153

c_{6} is correct to 12 digits.

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