Question

If \[{\left( {\dfrac{9}{4}} \right)^{ - 4}} \times {\left( {\dfrac{2}{3}} \right)^3} = {\left( {\dfrac{p}{q}} \right)^{11}}\], find the value of \[{\left( {\dfrac{p}{q}} \right)^{ - 2}}\]

Answer 1

Hint: Here the question is related to the exponents. The fractional numbers having the exponents. First we simplify the given equation and hence we determine the value of \[\left( {\dfrac{p}{q}} \right)\] and then we determine the \[{\left( {\dfrac{p}{q}} \right)^{ - 2}}\], so by using the law of exponents we simplify the given question.

Complete step by step solution:
The number which is present in the question is in the form of fraction. In decimal number we have a numerator and denominator. The exponential number is defined as a number of times the number is multiplied by itself.
Now consider the given question.
\[{\left( {\dfrac{9}{4}} \right)^{ - 4}} \times {\left( {\dfrac{2}{3}} \right)^3} = {\left( {\dfrac{p}{q}} \right)^{11}}\]
As we known by the law of exponents \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\] or if the fraction number has negative power means we can interchange the numerator value to the denominator and vice-versa. So we have
\[ \Rightarrow {\left( {\dfrac{4}{9}} \right)^4} \times {\left( {\dfrac{2}{3}} \right)^3} = {\left( {\dfrac{p}{q}} \right)^{11}}\]
The number 4 is a square of 2 and the number 9 is a square of 3. So the above inequality is written as
\[ \Rightarrow {\left( {\dfrac{{{2^2}}}{{{3^2}}}} \right)^4} \times {\left( {\dfrac{2}{3}} \right)^3} = {\left( {\dfrac{p}{q}} \right)^{11}}\]
The first term of the above inequality, the both numerator and denominator has power 2 we can consider as it whole. So we have
\[ \Rightarrow {\left( {{{\left( {\dfrac{2}{3}} \right)}^2}} \right)^4} \times {\left( {\dfrac{2}{3}} \right)^3} = {\left( {\dfrac{p}{q}} \right)^{11}}\]
By the law of exponents \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], the above inequality is written as
\[ \Rightarrow {\left( {\dfrac{2}{3}} \right)^8} \times {\left( {\dfrac{2}{3}} \right)^3} = {\left( {\dfrac{p}{q}} \right)^{11}}\]
By the law of exponents \[{a^m} \times {a^n} = {a^{m + n}}\], the above inequality is written as
\[ \Rightarrow {\left( {\dfrac{2}{3}} \right)^{8 + 3}} = {\left( {\dfrac{p}{q}} \right)^{11}}\]
On adding 8 and 3 we have
\[ \Rightarrow {\left( {\dfrac{2}{3}} \right)^{11}} = {\left( {\dfrac{p}{q}} \right)^{11}}\]
The power of the both exponents which is present on the LHS and RHS is the same. So we can equate the base value.
\[ \Rightarrow \dfrac{p}{q} = \dfrac{2}{3}\]
Hence we have determined the value of \[\left( {\dfrac{p}{q}} \right)\].
Now we determine the value of \[{\left( {\dfrac{p}{q}} \right)^{ - 2}}\]
So we have
\[ \Rightarrow {\left( {\dfrac{p}{q}} \right)^{ - 2}} = {\left( {\dfrac{2}{3}} \right)^{ - 2}}\]
As we known by the law of exponents \[{a^{ - m}} = \dfrac{1}{{{a^m}}}\] or if the fraction number has negative power means we can interchange the numerator value to the denominator and vice-versa. So we have
\[ \Rightarrow {\left( {\dfrac{p}{q}} \right)^{ - 2}} = {\left( {\dfrac{3}{2}} \right)^2}\]
On squaring both numerator and denominator we have
\[ \Rightarrow {\left( {\dfrac{p}{q}} \right)^{ - 2}} = \dfrac{9}{4}\]
Hence we have determined the value.
So, the correct answer is “\[\dfrac{9}{4}\]”.

Note: In the other numbers except fractions if the number has a negative power we will write the form of a fraction. If the fraction number has a negative power or exponent we can interchange the value of the numerator to the denominator and vice versa. This process is implemented to other kinds of numbers also. Because the value of the denominator will be considered as 1.