How do you evaluate $e^(-ln2) + lnsqrt(e^5) + 4e^(-2ln2)$ ?

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Answer 1 · 2017-02-21T07:46:52

$e^(-ln2)+lnsqrt(e^5)+4e^(-2ln2)=4$

As $e^a=b=>lnb=a$ .

Now putting $a=lnb$ in $e^a=b$ , we get $e^(lnb)=b$

Hence, $e^(-ln2)=1/e^(ln2)=1/2$ ,

$lnsqrt(e^5)=ln(e^5)^(1/2)=lne^(5/2)=5/2$ and

$4e^(-2ln2)=4e^(-ln2^2)=4e^(-ln4)=4xx1/e^(ln4)=4xx1/4=1$

Hence $e^(-ln2)+lnsqrt(e^5)+4e^(-2ln2)=1/2+5/2+1=4$

How do you evaluate e^(-ln2) + lnsqrt(e^5) + 4e^(-2ln2)e^(-ln2) + lnsqrt(e^5) + 4e^(-2ln2)?