Question 444093
ln &#8730 e
={{{lne^(1/2)}}}
={{{(1/2)lne}}}
={{{1/2}}}
since ln e = 1.
So ln root e =1/2

Log [&#8730 2 (4)]  right?
=log &#8730 2 +log 4
=log 2^(1/2)+log 4
=log 2^(1/2)+log 2^2
=log [2^(1/2)*2^2]
=log 2^(2.5)

You can use a calculator for that right?  Since it calculates to around .49 .  Hope that helps.