Question

Simplify \[{\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right]\]

Answer 1

Hint: Here, we are required to simplify the given exponential numbers. We will use the properties of exponents and convert the given numerators and denominators in their exponential form. We will then cancel out the denominator of the exponents of the terms. We will solve the expression further to get the required answer.

Formula Used:
We will use the formula \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\].

Complete step-by-step answer:
First, we will write the factors of the numbers in the numerators and denominators in their exponential form. Therefore, we get,
\[{\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right] = {\left( {\dfrac{{{3^4}}}{{{2^4}}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{{5^2}}}{{{3^2}}}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right]\]
Now, we know that \[\dfrac{{{a^m}}}{{{b^m}}} = {\left( {\dfrac{a}{b}} \right)^m}\].
Using this formula in the above equation, we get,
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right] = {\left( {\dfrac{3}{2}} \right)^{\dfrac{{ - 3 \times 4}}{4}}} \times \left[ {{{\left( {\dfrac{5}{3}} \right)}^{\dfrac{{ - 3 \times 2}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right]\]
Simplifying the exponent, we get
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right] = {\left( {\dfrac{3}{2}} \right)^{ - 3}} \times \left[ {{{\left( {\dfrac{5}{3}} \right)}^{ - 3}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right]\]
Now, since the power is negative, hence, to make it positive, we will do the inverse of the fractions. Therefore, we get
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right] = {\left( {\dfrac{2}{3}} \right)^3} \times \left[ {{{\left( {\dfrac{3}{5}} \right)}^3} \div {{\left( {\dfrac{2}{5}} \right)}^3}} \right]\]
Now, we know that multiplication is the reverse of division and vice-versa. Hence, to remove the division sign from the bracket, we will substitute it with a multiplication sign and do the reverse of the fraction which is acting as a divisor.
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right] = {\left( {\dfrac{2}{3}} \right)^3} \times \left[ {{{\left( {\dfrac{3}{5}} \right)}^3} \times {{\left( {\dfrac{5}{2}} \right)}^3}} \right]\]
Again, using the formula \[{\left( {\dfrac{a}{b}} \right)^m} = \dfrac{{{a^m}}}{{{b^m}}}\] and multiplying all the three fractions by removing the bracket between them, we get,
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right] = \dfrac{{{2^3}}}{{{3^3}}} \times \dfrac{{{3^3}}}{{{5^3}}} \times \dfrac{{{5^3}}}{{{2^3}}}\]
Now, cancelling out the same terms from the numerator and denominator, we get
\[ \Rightarrow {\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right] = 1\]
Therefore, \[{\left( {\dfrac{{81}}{{16}}} \right)^{\dfrac{{ - 3}}{4}}} \times \left[ {{{\left( {\dfrac{{25}}{9}} \right)}^{\dfrac{{ - 3}}{2}}} \div {{\left( {\dfrac{5}{2}} \right)}^{ - 3}}} \right]\] is simplified as 1.
Hence, this is the required answer.

Note: An exponent tells us how many times we have to multiply the base with itself to find the required number. Hence, when we were converting the numbers in their exponential form, we should take care that we have multiplied the correct number of factors and then written them in exponential form by counting them correctly. This is the most important step as if this step is wrong then solving this question will not be an easy task due to the presence of denominators in the given powers.