Question

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

Answer 1

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

Explanation:

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`