How do you evaluate #g(x)=lnx# for #x=e^-2#?

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Answer 1 · 2018-03-23T14:18:17

#g(x)=-2#

#g(x)=lne^(-2)---(1)#

using teh law of logs

#log_ab^n=nlog_ab#

#(1)rarr-2lne--(2)#

using the law

#log_a a=1#

we have

#lne=1#

#:.(2)rarr=-2#