Question

What is the simplified form of the expression, \[{{27}^{\dfrac{2}{3}}}\]?

Answer 1

Hint: Assume the value of the given expression as ‘E’. Use the properties of the topic ‘exponents and powers’ to simplify the given expression. Use the formula: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], where ‘a’ is called the base and ‘m’ and ‘n’ are called powers.

Complete step by step answer:
We have been provided to simplify the expression, \[{{27}^{\dfrac{2}{3}}}\]. Let us assume its simplified value as ‘E’.
\[\Rightarrow E={{27}^{\dfrac{2}{3}}}\]
This type of problems are asked in the topic ‘exponents and powers’ where we have to use various formulas and properties that help in simplification of a expression which may seem to be difficult in solving. The expression of exponents and powers are generally represented by \[{{a}^{m}}\], where ‘a’ is called the base and ‘m’ is called the power.
Now, let us come to the question. We have,
\[\Rightarrow E={{27}^{\dfrac{2}{3}}}\] - (i)
Here, we know that in exponential form 27 can be written as \[{{3}^{3}}\]. So, mathematically,
\[\Rightarrow 27={{3}^{3}}\]
Therefore, substituting the value of \[27={{3}^{3}}\] in expression (i), we get,
\[\Rightarrow E={{\left( {{3}^{3}} \right)}^{\dfrac{2}{3}}}\]
We know that one of the property of exponents are: - \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], therefore apply this formula, we have,
\[\Rightarrow E={{3}^{3\times \dfrac{2}{3}}}\]
Cancelling 3 in the numerator with 3 in the denominator, we get,
\[\begin{align}
  & \Rightarrow E={{3}^{2}} \\
 & \Rightarrow E=3\times 3 \\
 & \Rightarrow E=9 \\
\end{align}\]

Hence, the value of the simplified expression is 9.

Note: One may note that there are other methods also to solve this question. One of the methods can be a logarithmic method which is generally the inverse method of ‘exponents and powers’. We can take log to the base 3 both sides and then use the property, \[{{\log }_{3}}{{a}^{m}}=m{{\log }_{3}}a\] to get the simplified version. But, it is suggested to use the method used in the solution above. This is because logarithms are used for complex problems and are studied in higher classes.