Answer:
f(0.1) = [tex]i \pi[/tex]
Step-by-step explanation:
Plug x = 0.1.
Therefore f(0.1) = ln (log (0.1)).
0.1 = 10^(-1), and by the definition of log, log 0.1 = -1.
Now f(0.1) = ln (-1).
By Euler's identity, e^(i pi) = -1.
So ln (-1) would be i pi.
f(0.1) = [tex]i \pi[/tex]
