Question

Question 43

In the following question, fill in the blanks to make the statements true.

${[{\left(\frac{2}{13}\right)}^{-6}÷{\left(\frac{2}{13}\right)}^3]}^3\times {\left(\frac{2}{13}\right)}^{-9}=$ ___

Answer 1

Using laws of exponents,
$a^m÷a^n=(a{)}^{m-n},a^m\times a^n=(a{)}^{m+n}and(a^m{)}^n=(a{)}^{mn}$

$\therefore {[{\left(\frac{2}{13}\right)}^{-6}÷{\left(\frac{2}{13}\right)}^3]}^3\times {\left(\frac{2}{13}\right)}^{-9}={\left[{\left(\frac{2}{13}\right)}^{-6-3}\right]}^3\times {\left(\frac{2}{13}\right)}^{-9}$

$={\left[{\left(\frac{2}{13}\right)}^{-9}\right]}^3\times {\left(\frac{2}{13}\right)}^{-9}$

$={\left(\frac{2}{13}\right)}^{-27}\times {\left(\frac{2}{13}\right)}^{-9}$

$={\left(\frac{2}{13}\right)}^{-27-9}$

$={\left(\frac{2}{13}\right)}^{-36}Hence,{[{\left(\frac{2}{13}\right)}^{-6}÷{\left(\frac{2}{13}\right)}^3]}^3\times {\left(\frac{2}{13}\right)}^{-9}={\left(\frac{2}{13}\right)}^{-36}$