How do you simplify #(9a^3b^4)^(1/2)#?

2 Answers
Jim H
Aug 30, 2016

#3b^2a^(3/2)# or #3ab^2sqrta# depending on what is meant by "simplify".

Explanation:

Exponents distribute across multiplication, so

#(9a^3b^4)^(1/2) = 9^(1/2)(a^3)^(1/2)(b^4)^(1/2)#

Now use #9^(1/2) = sqrt9 = 3#.

Also use #(x^a)^b = x^(ab)#, to get

#(a^3)^(1/2) = a^(3/2)# which can be written #a*a^(1/2) = a sqrta#

and #(b^4)^(1/2) = b^2#.

#(9a^3b^4)^(1/2) = 9^(1/2)(a^3)^(1/2)(b^4)^(1/2)#

# = 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#.

Jim G.
Aug 30, 2016

#3a^(3/2)b^2#

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#

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