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

1 Answer
Aug 20, 2016

Answer:

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

Explanation:

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#.