Question

Fill in the blank to make the statement true.

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

Answer 1

Given that,

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

Step 1: Apply the formula $a^m÷a^n$ $or$ $\frac{a^m}{a^n}=a^{m-n}$ inside the brackets , we get

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

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

Step 2: Apply the formula ${\left(a^m\right)}^n=a^{mn}$ , we get

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

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

Step3: Apply the law of exponent $a^m\times a^n=a^{m+n}$ , we get

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

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

$=$${\left(\frac{2}{13}\right)}^{-36}$

Therefore,

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

Hence, the answer for the given blank is ${\left(\frac{2}{13}\right)}^{-36}$ .