Question

Solve: \[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}}\]

Answer 1

Hint: Given is the problem that is based on laws of indices and powers. We can see that the lowest base is 3. So we will convert 9 and 27 in the form of power of 3. Then applying the rules or laws of indices we will solve the problem.

Complete step by step answer:
Given that, \[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}}\]
As we know that 9 is the second power of 3 and 27 is the third power of 3. So we can write,
\[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}} = \dfrac{{{{\left( {{3^2}} \right)}^{\dfrac{1}{3}}} \times {{\left( {{3^3}} \right)}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}}\]
Now we know that, \[{\left( {{a^m}} \right)^n} = {a^{m \times n}}\]
\[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}} = \dfrac{{{{\left( 3 \right)}^{\dfrac{2}{3}}} \times {{\left( 3 \right)}^{\dfrac{{ - 3}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}}\]

Now we know that if the bases are same we can perform the respective operations directly on powers,
\[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}} =\dfrac{{{3^{\dfrac{2}{3} + \left( {\dfrac{{ - 3}}{2}} \right)}}}}{{{3^{\dfrac{{ - 1}}{6} + \left( {\dfrac{{ - 2}}{3}} \right)}}}}\]
\[\Rightarrow \dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}} = \dfrac{{{3^{\dfrac{{4 - 9}}{6}}}}}{{{3^{\dfrac{{ - 1 - 4}}{6}}}}}\]
\[\Rightarrow \dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}} = \dfrac{{{3^{\dfrac{{ - 5}}{6}}}}}{{{3^{\dfrac{{ - 5}}{6}}}}}\]
Since the numerator and denominator both are same we can cancel then directly,
\[\therefore \dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}} = \dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}} = 1\]

Therefore, the value of \[\dfrac{{{9^{\dfrac{1}{3}}} \times {{27}^{\dfrac{{ - 1}}{2}}}}}{{{3^{\dfrac{{ - 1}}{6}}} \times {3^{\dfrac{{ - 2}}{3}}}}}\] is $1$.

Note: When we solve the problem we should definitely get an idea of converting the bases into one same base since the other two bases are the powers of the smallest base coincidently. Then also note that we can add or subtract or perform the operation on power only and only if the bases are the same and are in product otherwise not. Also note that problems related to powers and bases are totally dependent on the laws of it.