How do you evaluate $3^ln0.2$ ?

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Answer 1 · 2017-02-24T17:34:56

$3^(ln0.2)=0.171$

Let $3^(ln0.2)=x$ . Taking natural log on both sides, we get

$ln0.2ln3=lnx$ and hence

$x=e^(ln0.2ln3)$

= $e^(-1.609xx1.0986)$

= $e^(-1.76815)=0.171$

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