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, $(e^0.24)^2=1.616$ .

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