How do you evaluate # (e^0.24)^2#?

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Answer 1 · 2016-08-20T07:45:59

#(e^0.24)^2=1.616#.

Let #x=(e^0.24)^2#

#:. x=e^((0.24)(2))=e^0.48#

To find its numerical value, let us use Log-table (base 10).

#x=e^0.48#

#rArr logx=loge^0.48=0.48loge#

#rArr logx=(0.48)(0.4343)=0.208464~=0.2085#

#rArr x=Antilog0.2085=1.616#

Hence, #(e^0.24)^2=1.616#.