Chapter ISG, Problem 7.3STP

To determine

To simplify each expression.

Answer to Problem 7.3STP

  ${27}^{\frac{2}{3}}-{16}^{\frac{3}{4}}$

  $=0$

  ${(\frac{4}{25})}^{-\frac{3}{2}}$

  $=\frac{125}{9}$

  ${(\frac{1}{4})}^{-2}\frac{{(x{y}^2)}^{-2}}{{(36x^6)}^{\frac{1}{2}}}$

  $=\frac{8}{3y^4{x}^{\frac{7}{2}}}$

Explanation of Solution

Given information:Equations given:

  ${27}^{\frac{2}{3}}-{16}^{\frac{3}{4}}$ =

  ${(\frac{4}{25})}^{-\frac{3}{2}}$ =

  ${(\frac{1}{4})}^{-2}\frac{{(x{y}^2)}^{-2}}{{(36x^6)}^{\frac{1}{2}}}$ =

Calculation:

On solving the given equation, ${27}^{\frac{2}{3}}-{16}^{\frac{3}{4}}$

27 can be written as $3^3$ and 16 can be written as $2^4$ .

Therefore,

  $={(3^3)}^{\frac{2}{3}}-{(2^4)}^{\frac{3}{4}}$

Opening the brackets.

  $=3^{\frac{3\times 2}{3}}-2^{\frac{4\times 3}{4}}$

On solving the equation,

  $=3^2-2^3$

Therefore,

  $=6-6$

So,

  $=0$

On solving the given equation, ${(\frac{4}{25})}^{-\frac{3}{2}}$

On solving further,

  $=\frac{{(4)}^{-\frac{3}{2}}}{{(25)}^{-\frac{3}{2}}}$ So,

4 can be written as $2^2$ and 25 can be written as $5^2$ .

  $=\frac{{(2^2)}^{-\frac{3}{2}}}{{(5^2)}^{-\frac{3}{2}}}$

Opening the brackets.

  $=\frac{2^{-\frac{2\times 3}{2}}}{5^{-\frac{2\times 3}{2}}}$

On solving further,

  $=\frac{2^{-3}}{5^{-3}}$

To remove the negative sign of power reverse the number.

  $=\frac{{(\frac{1}{2})}^3}{{(\frac{1}{5})}^3}$

On simplifying,

  $=\frac{5^3}{2^3}$

So ,

  $=\frac{125}{9}$

On solving the given equation, ${(\frac{1}{4})}^{-2}\frac{{(x{y}^2)}^{-2}}{{(36x^6)}^{\frac{1}{2}}}$

To remove the negative sign of power reverse the number.

36 can be written as ${(6)}^2$ .

  $={(4)}^2\frac{{(x{y}^2)}^{-2}}{{({(6)}^2{x}^6)}^{\frac{1}{2}}}$

To remove the negative sign of power reverse the number.

  $={(4)}^2\frac{{(\frac{1}{x{y}^2})}^2}{{(6)}^{\frac{2}{2}}{(x)}^{\frac{6\times 1}{2}}}$

Opening the brackets.

  $={(4)}^2\frac{\frac{1}{x^2{y}^4}}{(6){(x)}^{\frac{3}{2}}}$

On simplifying,

  $=\frac{{(4)}^2}{x^2{y}^{4}6{(x)}^{\frac{3}{2}}}$

Adding the powers,

  $=\frac{16}{y^4{(6)}^3{(x)}^{2+\frac{3}{2}}}$

On simplifying,

  $=\frac{16}{6y^4{(x)}^{\frac{4+3}{2}}}$

So,

  $=\frac{16}{6y^4{x}^{\frac{7}{2}}}$

Simplifying further,

  $=\frac{8}{3y^4{x}^{\frac{7}{2}}}$