Question

Solve the following terms \[{(81)^{ - 1}} \times {3^{ - 5}} \times {3^9} \times {(64)^{\dfrac{5}{6}}} \times {(\sqrt[3]{3})^6}\]

Answer 1

Hint: Indices: The base \[x\] raised to the power of \[p\] is equal to the multiplication of\[x\] p times.
\[{x^p} = x \times x \times x \times x \ldots \ldots .p{\rm{ }}times\], where x is the base and p is the indices.

Formula used:
Power rule:
\({({x^m})^n} = {x^{mn}}\)
\[{x^{{m^n}}} = {x^{({m^n})}}\]
To derive the solution of the given problem, we first need to solve each and every part of it. For that we have to use the indices formula and make the base as smallest as possible .Then multiply all the parts to find the value of the given problem.

Complete step by step answer:
We have to find out, \[{(81)^{ - 1}} \times {3^{ - 5}} \times {3^9} \times {(64)^{\dfrac{5}{6}}} \times {(\sqrt[3]{3})^6}\]
We will solve the problem part by part,
Let us consider the first part
The first part in the given problem is $(81)$\[^{ - 1}\]
Let us solve it, we get,
\[\Rightarrow {(81)^{ - 1}} = \dfrac{1}{{81}}\]
Let us write the denominator in terms of powers we get,
\[\Rightarrow \dfrac{1}{{81}} = \dfrac{1}{{{3^4}}}\]
\[\Rightarrow \dfrac{1}{{{3^4}}} = {({3^4})^{ - 1}}\]
Let us use the power rule in the above term,
\[\Rightarrow {({3^4})^{ - 1}} = {3^{ - 4}}\]
We have got\[{(81)^{ - 1}} = {3^{ - 4}}\].
Let us consider the second term,
The second term is \[{(64)^{\dfrac{5}{6}}}\]
Let us express the number in terms of power we get,
 \[\Rightarrow {(64)^{\dfrac{5}{6}}} = {({2^6})^{\dfrac{5}{6}}}\]
By using the power rule in the above equation we get,
\[\Rightarrow {({2^6})^{\dfrac{5}{6}}} = {2^{6 \times \dfrac{5}{6}}}\]
On solving the above equation we get,
\[\Rightarrow {2^{6 \times \dfrac{5}{6}}} = {2^5}\]
We have got,\[{(64)^{\dfrac{5}{6}}} = {2^5}\]
Now let us consider another term\[{(\sqrt[3]{3})^6}\]
Let us express the root in terms of power we get,
 \[\Rightarrow {(\sqrt[3]{3})^6} = {({3^{\dfrac{1}{3}}})^6}\]
Using the power rule in the above expression we get,
\[\Rightarrow {({3^{\dfrac{1}{3}}})^6} = {3^{\dfrac{1}{3} \times 6}}\]
And on further solving we get,
\[ {3^{\dfrac{1}{3} \times 6}} = {3^2}\]
Hence we have got,\[{(\sqrt[3]{3})^6} = {3^2}\]
Here we have expressed every term as a power of some base.
Therefore, by multiplying all the parts we have,
\[\Rightarrow {(81)^{ - 1}} \times {3^{ - 5}} \times {3^9} \times {(64)^{\dfrac{5}{6}}} \times {(\sqrt[3]{3})^6}\]\[ = {3^{ - 4}} \times {3^{ - 5}} \times {3^9} \times {2^5} \times {3^2}\]
We know that if the bases are same we can add the power using that identity we have,
\[{3^{ - 4}} \times {3^{ - 5}} \times {3^9} \times {2^5} \times {3^2}\]\[ = {3^{ - 4 + ( - 5) + 9 + 2}} \times {2^5}\]
And on further solving we get,
\[ {3^2} \times 32\]
\[= 9 \times 32\]
\[\Rightarrow {(81)^{ - 1}} \times {3^{ - 5}} \times {3^9} \times {(64)^{\dfrac{5}{6}}} \times {(\sqrt[3]{3})^6}\]\[ = 288\]

$\therefore$ We have found that \[{(81)^{ - 1}} \times {3^{-5}} \times {3^9} \times {(64)^{\dfrac{5}{6}}} \times {(\sqrt[3]{3})^6} = 288\].

Note:
We have used the following power values in the calculation\[\;81 = 3\;64 = 2\;\sqrt[3]{3} = {3^{\dfrac{1}{3}}}\] .
Also by expressing every as a power of some smaller term we have arrived at the result easily.