If k=64, how do you evaluate #1/2k^(2/3)+5(k^(1/2))^(2/3)#?
1 Answer
Dec 9, 2016
Explanation:
Using the
#color(blue)"laws of exponents"#
#color(red)(bar(ul(|color(white)(2/2)color(black)(a^(m/n)=(root(n)(a))^m" and "(a^m)^n=a^(mxxn))color(white)(2/2)|)))#
#rArr1/2k^(2/3)+5(k^(1/2))^(2/3)#
#=1/2k^(2/3)+5k^(1/2xx2/3)#
#=1/2k^(2/3)+5k^(1/3)# [let k = 64]
#=(1/2xx(root(3)(64))^2)+(5xxroot(3)(64))#
#=(1/2xx4^2)+(5xx4)#
#=8+20=28#
