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

$lne=1$

$:.(2)rarr=-2$

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