# How to solve (49x^2)^3/2 ÷ (9y^4)^3/2?

Jun 22, 2018

$= {\left(\frac{7 x}{3 {y}^{2}}\right)}^{3} = \frac{343 {x}^{3}}{27 {y}^{6}}$

#### Explanation:

Considering the question to be:

(49x^2)^(3/2) ÷ (9y^4)^(3/2)

Solution:
(49x^2)^(3/2) ÷ (9y^4)^(3/2) can be written as:

 =>(7^2 x^2)(3/2) ÷ (3^2y^4)^(3/2)

=> ((7x)^2)3/2 ÷( (3y^2)^2)^(3/2)

=>(7x)^(2xx3/2 )÷ ( 3y^2)^(2xx^3/2)

=>(7x)^3÷ ( 3y^2)^3 = ((7x)/ (3y^2))^3

$\implies \frac{343 {x}^{3}}{27 {y}^{6}}$

Jun 23, 2018

$= \frac{343}{27 {x}^{3}}$

#### Explanation:

${\left(49 {x}^{2}\right)}^{\frac{3}{2}} \setminus \div {\left(9 {y}^{4}\right)}^{\frac{3}{2}}$

Fractional exponent: the denominator is the value of the root, and the numerator is an exponent of the value (it goes inside the root).
So ${a}^{\frac{3}{2}}$ becomes $\setminus \sqrt{{a}^{3}}$

Thus, you now have
$\setminus \sqrt{{\left(49 {x}^{2}\right)}^{3}} \setminus \div \setminus \sqrt{{\left(9 {y}^{4}\right)}^{3}}$

Exponent of exponent: multiply them.
$\setminus \therefore \setminus \sqrt{117649 {x}^{6}} \setminus \div \setminus \sqrt{729 {x}^{12}}$

Change back to fractional exponents for the variable, so you can multiply:
$\left[\setminus \sqrt{117649} \setminus \cdot {\left({x}^{6}\right)}^{\frac{1}{2}}\right] \setminus \div \left[\setminus \sqrt{729} \setminus \cdot {\left({y}^{12}\right)}^{\frac{1}{2}}\right]$
$\left[\setminus \sqrt{117649} \setminus \cdot \left({x}^{3}\right)\right] \setminus \div \left[\setminus \sqrt{729} \setminus \cdot \left({y}^{6}\right)\right]$

Now simplify.
$\frac{\setminus \sqrt{117649} \setminus \cdot \left({x}^{3}\right)}{\setminus \sqrt{729} \setminus \cdot \left({y}^{6}\right)} = \frac{343 {x}^{3}}{27 {y}^{6}}$

(by the way, use a calculator for the square roots)