Question

How do you simplify \[{{32}^{-\dfrac{2}{5}}}\] ?

Answer 1

Hint: In the given question we have been asked to find the value of \[{{32}^{-\dfrac{2}{5}}}\] . In order to solve this question, first we need to factorize the number 32 and rewrite it as \[{{2}^{5}}\] , then using the laws of exponents and powers to solve the given question i.e. \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\] and \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\] . Using this property we solve the given question and simplify the result and we will get our required solution.
Formula used:
Using the laws of exponents which states that;
If ‘a’ is the rational number and ‘m’ and ‘n’ are the given rational exponent either positive exponent or negative exponent, then
 \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
If ‘a’ is the rational number and ‘m’ is the given rational negative exponent, then
 \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\]

Complete step-by-step answer:
We have given that,
 \[\Rightarrow {{32}^{-\dfrac{2}{5}}}\]
It is the exponential term.
Base = \[32\]
It can be expressed as;
 \[\Rightarrow 32=2\times 2\times 2\times 2\times 2\]
 \[\Rightarrow 32={{2}^{5}}\]
Therefore,
We can write the given term in the question as,
 \[\Rightarrow {{32}^{-\dfrac{2}{5}}}={{\left( {{2}^{5}} \right)}^{-\dfrac{2}{5}}}\]
Using the properties of laws and exponents that is given by,
 \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
Applying this property, we will obtain
 \[\Rightarrow {{32}^{-\dfrac{2}{5}}}={{\left( {{2}^{5}} \right)}^{-\dfrac{2}{5}}}={{2}^{5\times \left( -\dfrac{2}{5} \right)}}\]
Simplifying the powers in the above term, we will get
 \[\Rightarrow {{32}^{-\dfrac{2}{5}}}={{\left( {{2}^{5}} \right)}^{-\dfrac{2}{5}}}={{2}^{5\times \left( -\dfrac{2}{5} \right)}}={{2}^{-2}}\]
Using the properties of laws and exponents for the negative exponent that is given by,
 \[{{a}^{-m}}=\dfrac{1}{{{a}^{m}}}\]
Applying this property, we will obtain
 \[\Rightarrow {{32}^{-\dfrac{2}{5}}}={{\left( {{2}^{5}} \right)}^{-\dfrac{2}{5}}}={{2}^{5\times \left( -\dfrac{2}{5} \right)}}={{2}^{-2}}=\dfrac{1}{{{2}^{2}}}\]
Simplifying the powers in the above term, we will get
 \[\Rightarrow {{32}^{-\dfrac{2}{5}}}={{\left( {{2}^{5}} \right)}^{-\dfrac{2}{5}}}={{2}^{5\times \left( -\dfrac{2}{5} \right)}}={{2}^{-2}}=\dfrac{1}{{{2}^{2}}}=\dfrac{1}{4}\]
Therefore,
 \[\Rightarrow {{32}^{-\dfrac{2}{5}}}=\dfrac{1}{4}\]
Hence, this is the required answer.
So, the correct answer is “$\dfrac{1}{4}$”.

Note: While solving these types of questions, students should remember all the laws of exponents as they will be able to solve the question easily. Students need to remember that also if a number is expressed in exponential form and has a negative exponent, then first we need to convert the negative exponent to a positive exponent by taking the reciprocal of the base. While applying any law of exponents we should very carefully write the terms in a respective form as it is given in the associated law to it.