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

1 Answer

Shwetank Mauria

Feb 24, 2017

#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#

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