Chapter 7.2, Problem 24P

Simplify each expression

$\left(7q^{\frac{4}{3}}\cdot 6r^{\frac{3}{5}}\right)\cdot \left(7q^{\frac{1}{3}}\cdot 6r^{\frac{1}{5}}\right)$

To determine

To simplify the expression $\left(7q^{\frac{4}{3}}.6r^{\frac{3}{5}}\right).\left(7q^{\frac{1}{3}}.6r^{\frac{1}{5}}\right)$

Answer to Problem 24P

The simplest form of the given expression $\left(7q^{\frac{4}{3}}.6r^{\frac{3}{5}}\right).\left(7q^{\frac{1}{3}}.6r^{\frac{1}{5}}\right)$ is $36h^{\left(\frac{15}{4}\right)}{k}^{\left(\frac{9}{4}\right)}$

Explanation of Solution

Given information :

Given expression $\left(7q^{\frac{4}{3}}.6r^{\frac{3}{5}}\right).\left(7q^{\frac{1}{3}}.6r^{\frac{1}{5}}\right)$

As we know that ${\left(a^m\right)}^n=a^{m\times n}\mathrm{........}\text{(i)}$

Now consider $\left(7q^{\frac{4}{3}}.6r^{\frac{3}{5}}\right).\left(7q^{\frac{1}{3}}.6r^{\frac{1}{5}}\right)=\left(7\times 7\right)\left(6\times 6\right)\left(q^{\frac{4}{3}}.q^{\frac{1}{3}}\right)\left(r^{\frac{3}{5}}.r^{\frac{1}{5}}\right)$

  $\begin{array}{l}=\left(49\right)\left(36\right){\left(q\right)}^{\frac{4}{3}+\frac{1}{3}}{\left(r\right)}^{\frac{3}{5}+\frac{1}{5}}\text{from}\text{rule}(i)\\ =1764q^{\left(\frac{4+1}{3}\right)}{r}^{\left(\frac{3+1}{5}\right)}\\ \therefore \left(7q^{\frac{4}{3}}.6r^{\frac{3}{5}}\right).\left(7q^{\frac{1}{3}}.6r^{\frac{1}{5}}\right)=1764q^{\left(\frac{5}{3}\right)}{r}^{\left(\frac{4}{5}\right)}\end{array}$

Hence, the simplest form of the given expression $\left(7q^{\frac{4}{3}}.6r^{\frac{3}{5}}\right).\left(7q^{\frac{1}{3}}.6r^{\frac{1}{5}}\right)$ is $1764q^{\left(\frac{4+1}{3}\right)}{r}^{\left(\frac{3+1}{5}\right)}$