Question

Which of the following is equivalent to ${\left(\frac{1}{3}\right)}^{-4}{\left(\frac{1}{9}\right)}^{-3}{\left(\frac{1}{27}\right)}^{-2}$?

• A. ${\left(\frac{1}{3}\right)}^{-8}$
• B. ${\left(\frac{1}{3}\right)}^{-9}$
• C. ${\left(\frac{1}{3}\right)}^{-16}$
• D. ${\left(\frac{1}{3}\right)}^{-18}$
• E. ${\left(\frac{1}{3}\right)}^{-144}$

Answer 1

The correct option is C ${\left(\frac{1}{3}\right)}^{-16}$

For the given question, we make use of the law of exponent:

$(\frac{1}{a}{)}^{-m}$= $\left(a^m\right)$

So, in the given problem:

$(\frac{1}{3}{)}^{-4}$= $\left(3^4\right)$

$(\frac{1}{9}{)}^{-3}$= $\left(9^3\right)=(3^2{)}^3=3^6$

$(\frac{1}{27}{)}^{-2}$= $\left({27}^2\right)=(3^3{)}^2=3^6$

$\therefore$

$3^4\times 3^6\times 3^6=3^{16}=(\frac{1}{3}{)}^{-16}$