How do you simplify ${(9 a^{3} b^{4})}^{\frac{1}{2}}$ $(9a^3b^4)^(1/2)$ ?

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How do you simplify ${(9 a^{3} b^{4})}^{\frac{1}{2}}$ $(9a^3b^4)^(1/2)$ ?

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Answer 1 · 2016-08-30T17:12:34

$3 b^{2} a^{\frac{3}{2}}$ $3b^2a^(3/2)$ or $3 a b^{2} \sqrt{a}$ $3ab^2sqrta$ depending on what is meant by "simplify".

Explanation:

Exponents distribute across multiplication, so

${(9 a^{3} b^{4})}^{\frac{1}{2}} = 9^{\frac{1}{2}} {(a^{3})}^{\frac{1}{2}} {(b^{4})}^{\frac{1}{2}}$ $(9a^3b^4)^(1/2) = 9^(1/2)(a^3)^(1/2)(b^4)^(1/2)$

Now use $9^{\frac{1}{2}} = \sqrt{9} = 3$ $9^(1/2) = sqrt9 = 3$ .

Also use ${(x^{a})}^{b} = x^{a b}$ $(x^a)^b = x^(ab)$ , to get

${(a^{3})}^{\frac{1}{2}} = a^{\frac{3}{2}}$ $(a^3)^(1/2) = a^(3/2)$ which can be written $a \cdot a^{\frac{1}{2}} = a \sqrt{a}$ $a*a^(1/2) = a sqrta$

and ${(b^{4})}^{\frac{1}{2}} = b^{2}$ $(b^4)^(1/2) = b^2$ .

${(9 a^{3} b^{4})}^{\frac{1}{2}} = 9^{\frac{1}{2}} {(a^{3})}^{\frac{1}{2}} {(b^{4})}^{\frac{1}{2}}$ $(9a^3b^4)^(1/2) = 9^(1/2)(a^3)^(1/2)(b^4)^(1/2)$

$= 3 a^{\frac{3}{2}} b^{2}$ $= 3a^(3/2)b^2$ $" "$ (which is simpler in some sense)

$= 3 a \sqrt{a} b^{2}$ $= 3asqrtab^2$

which is not as easy to read as putting the radical last

$= 3 a b^{2} \sqrt{a}$ $= 3ab^2sqrta$ .

Answer 2 · 2016-08-30T17:17:32

$3 a^{\frac{3}{2}} b^{2}$ $3a^(3/2)b^2$

Explanation:

Using the $laws of exponents$ $color(blue)"laws of exponents"$

$color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)((a^m)^n=a^(mn))color(white)(a/a)|)))$

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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