What is (2x^{1/2}y^{1/3})(3x^{1/2}y) ?

The solution to (2x^{1/2}y^{1/3})(3x^{1/2}y) is 6xy^{4/3}

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simplify (2x^{1/2}y^{1/3})(3x^{1/2}y)

simplify

(2 x^{\frac{1}{2}} y^{\frac{1}{3}}) (3 x^{\frac{1}{2}} y)

6 x y^{\frac{4}{3}}

Solution steps

(2 x^{\frac{1}{2}} y^{\frac{1}{3}}) (3 x^{\frac{1}{2}} y)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

x^{\frac{1}{2}} x^{\frac{1}{2}} = x^{\frac{1}{2} + \frac{1}{2}}

= 2 y^{\frac{1}{3}} \cdot 3 x^{\frac{1}{2} + \frac{1}{2}} y

Combine the fractions

\frac{1}{2} + \frac{1}{2} : 1

= 2 \cdot 3 x^{1} y y^{\frac{1}{3}}

Apply rule

a^{1} = a

x^{1} = x

= 2 \cdot 3 x y y^{\frac{1}{3}}

Multiply the numbers:

2 \cdot 3 = 6

= 6 x y y^{\frac{1}{3}}

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

y^{\frac{1}{3}} y = y^{\frac{1}{3} + 1}

= 6 x y^{\frac{1}{3} + 1}

y^{\frac{1}{3} + 1} = y^{\frac{4}{3}}

= 6 x y^{\frac{4}{3}}

(\frac{1}{2} p q^{2}) (\frac{1}{2} p^{2} q)

5 y y y y

(22 x^{4}) (5 x^{5})

e^{- X}

What is (2x^{1/2}y^{1/3})(3x^{1/2}y) ?
The solution to (2x^{1/2}y^{1/3})(3x^{1/2}y) is 6xy^{4/3}