Question

How do you evaluate $ {{f}^{-2}}g $ for $ f=3 $ and $ g=27 $ ?

Answer 1

Hint: We use the indices theorem of \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\]. We change the equation $ {{f}^{-2}}g $ to the form of positive indices. We then put the values $ f=3 $ and $ g=27 $ to find the value of $ {{f}^{-2}}g $ .

Complete step by step answer:
We have two variables and their values are $ f=3 $ and $ g=27 $ .
We need to find the value of $ {{f}^{-2}}g $ . This is the multiplication of $ {{f}^{-2}} $ and $ g $ .
We know the theorem of indices for \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\].
Using the theorem for the variable f we get \[{{f}^{-2}}=\dfrac{1}{{{f}^{2}}}\].
Therefore, $ {{f}^{-2}}g=\dfrac{g}{{{f}^{2}}} $ .
We place the values in the equation and get $ {{f}^{-2}}g=\dfrac{g}{{{f}^{2}}}=\dfrac{27}{{{3}^{2}}}=\dfrac{27}{9}=3 $ .
The simplified value of $ {{f}^{-2}}g $ is 3.

Note:
 The values all are powered values of 3. Therefore, we can convert in the form of 3. We have the value of g as 27. We know $ g=27={{3}^{3}} $ .
We have indices theorem for \[{{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}}\] and \[{{a}^{x}}{{a}^{y}}={{a}^{x+y}}\].
Therefore, we get $ {{f}^{-2}}g={{3}^{-2}}\times {{3}^{3}}={{3}^{3-2}}=3 $ .