Question

Find the product.

(i) $a^2\times \left(2a^{22}\right)\times \left(4a^{26}\right)$

(ii) $\left(\frac{2}{3}xy\right)\times (-\frac{9}{10}{x}^2{y}^2)$

(iii) $\left(\begin{array}{c}\frac{-10}{3}p{q}^3\end{array}\right)\times \left(\begin{array}{c}\frac{6}{5}{p}^{3}q\end{array}\right)$

(iv) $x\times x^2\times x\times x^3\times x^4$

Answer 1

As we know that when the exponential terms get multiplied with the same bases, then we can add their powers as
i.e., $a^m\times a^n=a^{(m+n)}$ ......(1)
This is also known as product law/rule.

(i) $a^2\times \left(2a^{22}\right)\times \left(4a^{26}\right)$

$=(2\times 4)\left(a^{2+22+26}\right)$ [from equation (1)]

$=8a^{2+22+26}=8a^{50}$

$\therefore a^2\times \left(2a^{22}\right)\times \left(4a^{26}\right)=8a^{50}$

(ii) $(\frac{2}{3}\times xy)\left(\frac{-9}{10}{x}^2{y}^2\right)$

$=\frac{2\times (-9)}{3\times 10}\times x^{(1+2)}\times y^{(1+2)}$ [from equation(1)]
$=\frac{-3}{5}{x}^3{y}^3$

$\therefore (\frac{2}{3}\times xy)\left(\frac{-9}{10}{x}^2{y}^2\right)=\frac{-3}{5}{x}^3{y}^3$

(iii) $\left(\frac{-10}{3}p{q}^3\right)\times \left(\frac{6}{5}{p}^{3}q\right)$

$=\frac{(-10)\times 6}{3\times 5}\times p^{1+3}\times q^{3+1}$ [from equation(1)]

$=-4p^4{q}^4$

$\therefore \left(\frac{-10}{3}p{q}^3\right)\times \left(\frac{6}{5}{p}^{3}q\right)=-4p^4{q}^4$

(iv) $x\times x^2\times x\times x^3\times x^4$

$=x^{1+2+1+3+4}$ [from equation(1)]

$=x^{11}$

$\therefore x\times x^2\times x\times x^3\times x^4=x^{11}$