Question

Simplify and express the result in power notation with positive exponent.
$\left( i \right){\left( { - 4} \right)^5} \div {\left( { - 4} \right)^8}$
$\left( {ii} \right){\left( {\dfrac{1}{{{2^3}}}} \right)^2}$
$\left( {iii} \right){\left( { - 3} \right)^4} \times {\left( {\dfrac{5}{3}} \right)^4}$
$\left( {iv} \right)\left( {{3^{ - 7}} \div {3^{ - 10}}} \right) \times {3^{ - 5}}$
$\left( v \right){2^{ - 3}} \times {\left( { - 7} \right)^{ - 3}}$

Answer 1

Hint – In this particular question use the concept that the resultant in power notation with positive extent is written as ${\left( a \right)^{ + n}}$ or ${\left( {\dfrac{1}{a}} \right)^{ + n}}$or -${\left( a \right)^{ + n}}$ or -${\left( {\dfrac{1}{a}} \right)^{ + n}}$ so use this concept to reach the solution of the question.

Complete step by step solution:
We have to simplify these expressions and we have to express the resultant in power notation with positive extent.
For example: we have to express in the terms of ${\left( a \right)^{ + n}}$ or ${\left( {\dfrac{1}{a}} \right)^{ + n}}$or -${\left( a \right)^{ + n}}$ or -${\left( {\dfrac{1}{a}} \right)^{ + n}}$.
Where, (a) and (n) are some positive constants.
$\left( i \right){\left( { - 4} \right)^5} \div {\left( { - 4} \right)^8}$
Now as we know that negative one to the power of odd numbers is always negative one.
And a negative one to the power of even numbers is always a positive one.
So first simplify the above equation according to the above property we have,
$ \Rightarrow \left[ {{{\left( { - 1} \right)}^5}{{\left( 4 \right)}^5}} \right] \div \left[ {{{\left( { - 1} \right)}^8}{{\left( 4 \right)}^8}} \right]$
$ \Rightarrow \left[ { - {{\left( 4 \right)}^5}} \right] \div \left[ { + {{\left( 4 \right)}^8}} \right]$
$ \Rightarrow - \dfrac{{{{\left( 4 \right)}^5}}}{{{4^8}}}$
Now use the property that ($\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}{\text{ or }} = \dfrac{1}{{{a^{n - m}}}}$), so we have
$ \Rightarrow - \dfrac{1}{{{4^{8 - 5}}}} = - \dfrac{1}{{{4^3}}}$
Now as we know that, $4 = {2^2}$ so we have,
$ \Rightarrow - \dfrac{1}{{{4^{8 - 5}}}} = - \dfrac{1}{{{4^3}}} = - \dfrac{1}{{{{\left( {{2^2}} \right)}^3}}} = - \dfrac{1}{{{2^6}}}$
So this is the required simplification of the first expression.
$\left( {ii} \right){\left( {\dfrac{1}{{{2^3}}}} \right)^2}$
As we know that the square of 1 is always 1.
Therefore, ${\left( {\dfrac{1}{{{2^3}}}} \right)^2} = \dfrac{1}{{{{\left( {{2^3}} \right)}^2}}} = \dfrac{1}{{{2^6}}}$
So this is the required simplification of the second expression.
$\left( {iii} \right){\left( { - 3} \right)^4} \times {\left( {\dfrac{5}{3}} \right)^4}$
So first simplify the above equation according to the above written property we have,
$ \Rightarrow {\left( { - 3} \right)^4} \times {\left( {\dfrac{5}{3}} \right)^4} = {\left( { - 1} \right)^4}{\left( 3 \right)^4} \times {\left( {\dfrac{5}{3}} \right)^4} = + {\left( 3 \right)^4} \times {\left( {\dfrac{5}{3}} \right)^4}$
Now again simplify we have,
\[ \Rightarrow {\left( 3 \right)^4} \times {\left( {\dfrac{5}{3}} \right)^4} = {\left( 3 \right)^4} \times \dfrac{{{5^4}}}{{{3^4}}} = {5^4}\]
So this is the required simplification of the third expression.
$\left( {iv} \right)\left( {{3^{ - 7}} \div {3^{ - 10}}} \right) \times {3^{ - 5}}$
Now in the above equation use the property that $\dfrac{{{a^{ - b}}}}{{{c^{ - d}}}} = \dfrac{{{c^d}}}{{{a^b}}}$ so we have,
  $ \Rightarrow \left( {{3^{ - 7}} \div {3^{ - 10}}} \right) \times {3^{ - 5}} = \left( {{3^{10}} \div {3^7}} \right) \times {3^{ - 5}}$
$ \Rightarrow \left( {{3^{10}} \div {3^7}} \right) \times {3^{ - 5}} = \left( {\dfrac{{{3^{10}}}}{{{3^7}}}} \right) \times {3^{ - 5}}$
Now use the property that ($\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}{\text{ or }} = \dfrac{1}{{{a^{n - m}}}}$), so we have
$ \Rightarrow \left( {\dfrac{{{3^{10}}}}{{{3^7}}}} \right) \times {3^{ - 5}} = \left( {{3^{10 - 7}}} \right) \times {3^{ - 5}} = {3^3} \times {3^{ - 5}} = {3^{3 - 5}} = {3^{ - 2}} = \dfrac{1}{{{3^2}}}$
So this is the required simplification of the fourth expression.
$\left( v \right){2^{ - 3}} \times {\left( { - 7} \right)^{ - 3}}$
Now simplify this we have,
$ \Rightarrow {2^{ - 3}} \times {\left( { - 7} \right)^{ - 3}} = \dfrac{1}{{{2^3} \times {{\left( { - 7} \right)}^3}}} = \dfrac{1}{{{2^3} \times {{\left( { - 1} \right)}^3}{{\left( 7 \right)}^3}}} = \dfrac{{ - 1}}{{{2^3} \times {{\left( 7 \right)}^3}}} = \dfrac{{ - 1}}{{{{\left( {2 \times 7} \right)}^3}}} = \dfrac{{ - 1}}{{{{\left( {14} \right)}^3}}}$
So this is the required simplification of the fifth expression.
So this is the required answer.

Note – Whenever we face such types of questions the key concept we have to remember is that always remember that a negative one to the power of an odd number is always a negative one. And negative one to the power of even numbers is always positive one so simplify accordingly as above, we will get the required answer.