Question

Evaluate the following expression: \[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}}\]

Answer 1

Hint: First we have to know what surds and indices. Mention their laws. Then convert the terms in the given expression as a square of any number (if possible). Then using laws of surds and indices evaluate the given expression.

Complete step by step solution:
Let \[a\] be a rational number and \[n\] be a positive integer such that \[{a^{\dfrac{1}{n}}} = \sqrt[n]{a}\] is irrational. Then, \[\sqrt[n]{a}\] is called \[a\] surd of order \[n\].
Laws of surds: Suppose \[a\] and \[b\] are any two non-zero real numbers. Let \[n\] and \[m\] be any two non-zero integers then the following laws holds:
(i) \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\]
(ii) \[\sqrt[n]{{ab}} = {a^{\dfrac{1}{n}}} \times {b^{\dfrac{1}{n}}}\]
(iii) \[\sqrt[n]{{\dfrac{a}{b}}} = \dfrac{{\sqrt[n]{a}}}{{\sqrt[n]{b}}}\]
(iv) \[\left( {\sqrt[n]{a}} \right) = a\]
(v) \[\sqrt[m]{{\left( {\sqrt[n]{a}} \right)}} = \sqrt[{mn}]{a}\]
(vi) \[{\left( {\sqrt[m]{a}} \right)^n} = \sqrt[m]{{\left( {{a^n}} \right)}}\].
Given \[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}}\]----(1)
Since we know that \[4 = {2^2}\]and \[9 = {3^2}\], then the expression (1) becomes
\[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}} = {\left( {{{\left( {\dfrac{2}{3}} \right)}^2}} \right)^{ - \dfrac{3}{2}}}\]---(2)
Using the fourth and fifth laws of surds in the equation (2), we get
\[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}} = {\left( {\dfrac{2}{3}} \right)^{ - \dfrac{3}{2} \times 2}}\] ---(3)
Simplifying the equation (3), we get
\[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}} = {\left( {\dfrac{2}{3}} \right)^{ - 3}}\]--(4)
Since we know that \[{x^{ - n}} = \dfrac{1}{{{x^n}}}\], then the equation (4), we get
\[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}} = \dfrac{1}{{{{\left( {\dfrac{2}{3}} \right)}^3}}}\]----(5)
Using the law of indices, the equation (4) becomes
 \[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}} = \dfrac{1}{{\dfrac{{{2^3}}}{{{3^3}}}}} = {\left( {\dfrac{3}{2}} \right)^3}\]---(6)
Simplifying the equation (6), we get
\[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}} = \dfrac{{27}}{8} = 3.375\]
Hence, the value of \[{\left( {\dfrac{4}{9}} \right)^{ - \dfrac{3}{2}}}\] is \[\dfrac{{27}}{8} = 3.375\].

Note:
Laws of indices: Suppose \[a\] and \[b\] are any two non-zero real numbers. Let \[n\] and \[m\] be any two non-zero integers then the following laws holds: