Question

Show that \[{(8{x^2})^{8 - r}}{\left( {\dfrac{1}{{2x}}} \right)^r} = {2^{24 - 4r}}({x^{16 - 3r}})\]

Answer 1

Hint: Here we have to prove the given term as equate. By using the power rule formula to find the LHS of the given data. On doing some simplification we get the required answer for RHS.
Indices: The base x raised to the power of p is equal to the multiplication of \[x,p\] times.
\[{x^p} = x \times x \times x \times x \ldots \ldots .p\] times, where, \[x\] is the base and \[p\] is the indices.

Formula Used:
Power rule:
1) \[{\left( {{x^m}} \right)^n} = {x^{mn}}\]
2) \[{x^{{m^n}}} = {x^{^{\left( {{m^n}} \right)}}}\]
To derive the solution of the given problem we need to show left hand side equals to right hand side. For that we first need to solve each and every part of it. We have to use the indices formula and make the base as small as possible .Then multiply all the parts to find the value of the given problem given in the right hand side.

Complete step by step answer:
We have to show that, \[{(8{x^2})^{8 - r}}{\left( {\dfrac{1}{{2x}}} \right)^r} = {2^{24 - 4r}}({x^{16 - 3r}})\]
We will solve first the parts of the problem then the whole problem from the left hand side.
The first part is \[{(8{x^2})^{8 - r}}\].
Solving we get, \[{(8{x^2})^{8 - r}} = {({2^3}{x^2})^{8 - r}}\] .
Applying the power rule \[{\left( {{x^m}} \right)^n} = {x^{mn}}\] we have,
\[ \Rightarrow {(8{x^2})^{8 - r}} = {2^3}^{\left( {8 - r} \right)}{x^{2\left( {8 - r} \right)}}\]
On multiply the term and we get,
\[ \Rightarrow {(8{x^2})^{8 - r}} = {2^{24 - 3r}}{x^{16 - 2r}}\]
Considering the second term,\[{\left( {\dfrac{1}{{2x}}} \right)^r}\]
 Solving we get,\[{\left( {\dfrac{1}{{2x}}} \right)^r} = {\left( {{2^{ - 1}}{x^{ - 1}}} \right)^r}\]
Applying the power rule \[{\left( {{x^m}} \right)^n} = {x^{mn}}\] we have,
\[{\left( {\dfrac{1}{{2x}}} \right)^r} = \left( {{2^{ - r}}{x^{ - r}}} \right)\]
Now from left hand side, multiplying the terms we get by simplifying we get,
L.H.S. = \[{(8{x^2})^{8 - r}}{\left( {\dfrac{1}{{2x}}} \right)^r}\]
\[ \Rightarrow {2^{24 - 3r}}{x^{16 - 2r}} \times \left( {{2^{ - r}}{x^{ - r}}} \right)\]
On multiplication we get,
\[ \Rightarrow {2^{24 - 3r - r}}{x^{16 - 2r - r}}\]
On simplify we get,
\[ \Rightarrow {2^{24 - 4r}}{x^{16 - 3r}}\]
R.H.S
Thus,\[{(8{x^2})^{8 - r}}{\left( {\dfrac{1}{{2x}}} \right)^r} = {2^{24 - 4r}}({x^{16 - 3r}})\]
Hence, we get Left hand side = Right hand side.

Note: An index number is a number which is raised to a power. The power, also known as the index, tells you how many times you have to multiply the number by itself. The plural form of index is indices.
For example, \[{2^5}\]means that you have to multiply \[2\] by itself five times =\[2 \times 2 \times 2 \times 2 \times 2 = 32\].
Here we have used the formula, \[\left( {\dfrac{1}{{{x^n}}}} \right) = {x^{ - n}}\]