How do you simplify #(9a^3b^4)^(1/2)#?
2 Answers
Explanation:
Exponents distribute across multiplication, so
Now use
Also use
and
# = 3a^(3/2)b^2# #" "# (which is simpler in some sense)
# = 3asqrtab^2#
which is not as easy to read as putting the radical last
# = 3ab^2sqrta# .
Explanation:
Using the
#color(blue)"laws of exponents"#
#color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)((a^m)^n=a^(mn))color(white)(a/a)|)))# This law applies to each value inside the bracket.
#rArr9^(1xx1/2xxa^(3xx1/2)xxb^(4xx1/2)=9^(1/2)a^(3/2)b^2#
#color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(a^(1/2)=sqrta)color(white)(a/a)|)))#
#rArr9^(1/2)=sqrt9=3#
#rArr(9a^3b^4)^(1/2)=3a^(3/2)b^2#