Question

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

Answer 1

`3^(ln0.2)=0.171`

Explanation:

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`