Use bisection method to find the roots of x+logx-2 which lies between 1 and 2
Answers
Answer:
l don't know this answer
Given : x+logx-2
To Find : root which lies between 1 and 2 Using bisection method
Solution:
f(x) = x + logx - 2
considering natural log :
f(1) = 1 + log (1) - 2 = - 1
f(2) = 2 + log (2) - 2 = log 2 = 0.693
as 0 lies between - 1 and 0.693
Hence root lies between 1 and 2
mid point of 1 and 2 is 1.5
f(1.5) = 1.5 + log(1.5) - 2 = -0.0945
0 lies between -0.0945 and 0.693
Hence root lies between
1.5 and 2
Continuing this process
x Value
1 -1.0
2 0.693
1.5 -0.095
1.75 0.310
1.625 0.111
1.5625 0.009
1.53125 -0.043
1.546875 -0.017
1.5546875 -0.004
1.55859375 0.002
1.556640625 -0.001
1.557617188 0.001
1.557128906 -0.00003
1.557373047 0.00037
Hence x ≈ 1.557
considering log to the base 10
x Value
1 -1.0
2 0.3
1.5 -0.32391
1.75 -0.00696
1.875 0.14800
1.8125 0.07078
1.78125 0.03197
1.765625 0.01252
1.7578125 0.00279
1.75390625 -0.00209
1.755859375 0.00035
1.754882813 -0.00087
1.755371094 -0.00026
1.755615234 0.00004
1.755493164 -0.00011
1.755554199 -0.00003
1.755584717 0.000007
1.755569458 -0.000013
1.755577087 -0.000003
x ≈ 1.7555
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