Question

How do you evaluate `g(x)=-lnx` for `x=1/2`?

Answer 1

For `x=1/2; g(x) ~~0.693147` ;#

Explanation:

`x= 1/2 , g(x) = - ln(x) :. g(1/2) = -ln(1/2) = -[ln(1)-ln(2)] ~~ - [0-0.693147) ~~0.693147`
For `x=1/2; g(x) ~~0.693147` ;#
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