Question

(a) Find the reciprocal of: ${{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}\div {{\left( \dfrac{3}{4} \right)}^{0}}$ .
(b) Simplify: ${{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}$ .

Answer 1

Hint: For the first problem we need to calculate the reciprocal of the given value. For this we will firs simplify the given expression by using the some exponential formulas like ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ and ${{a}^{0}}=1$ . We will simplify the given expression by using the above formula and use the basic math rule which says that the division of any value with $1$ is equal to that value. So after having the simplified value of the given expression we will calculate the reciprocal by using the formula ${{a}^{-1}}=\dfrac{1}{a}$ .
For the second problem we need to simplify the given expression. We can observe that all the given values are having the power of $-1$ which represents the reciprocal values. We will use the exponential formula ${{a}^{-1}}=\dfrac{1}{a}$ and convert all of them into fractional form. Now we will simplify each value by using the basic mathematical operations.

Complete step-by-step solution:
a). Given expression is ${{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}\div {{\left( \dfrac{3}{4} \right)}^{0}}$.
Considering the value ${{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}$ . applying the exponential formula ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ in the above value, then we will get
$\begin{align}
  & {{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}={{\left( \dfrac{3}{4} \right)}^{-2+3}} \\
 & \Rightarrow {{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}={{\left( \dfrac{3}{4} \right)}^{1}} \\
 & \Rightarrow {{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}=\dfrac{3}{4} \\
\end{align}$
Considering the value ${{\left( \dfrac{3}{4} \right)}^{0}}$. Applying the exponential formula ${{a}^{0}}=1$, then we will have
${{\left( \dfrac{3}{4} \right)}^{0}}=1$
Substituting all the values we have in the given expression, then we will get
${{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}\div {{\left( \dfrac{3}{4} \right)}^{0}}=\dfrac{3}{4}\div 1$
When we are dividing a value with $1$, then we will get the same value as a result. So the value of above expression will be
${{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}\div {{\left( \dfrac{3}{4} \right)}^{0}}=\dfrac{3}{4}$
Now the reciprocal of the above value is given by
$\begin{align}
  & {{\left( \dfrac{3}{4} \right)}^{-1}}=\dfrac{1}{\dfrac{3}{4}} \\
 & \Rightarrow {{\left( \dfrac{3}{4} \right)}^{-1}}=\dfrac{4}{3} \\
\end{align}$
Hence the reciprocal of the given expression ${{\left( \dfrac{3}{4} \right)}^{-2}}\times {{\left( \dfrac{3}{4} \right)}^{3}}\div {{\left( \dfrac{3}{4} \right)}^{0}}$ is $\dfrac{4}{3}$ .

(b)
Given expression is ${{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}$.
We can observe that all the values in the above expression have a power of $-1$. So using the exponential formula ${{a}^{-1}}=\dfrac{1}{a}$ and simplifying the expression by using the basic mathematical operations, then we will have
$\begin{align}
  & {{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}={{\left( \dfrac{1}{2}-\dfrac{1}{3} \right)}^{-1}}+{{\left( \dfrac{1}{6}-\dfrac{1}{8} \right)}^{-1}} \\
 & \Rightarrow {{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}={{\left( \dfrac{3\times 1-2\times 1}{2\times 3} \right)}^{-1}}+{{\left( \dfrac{8\times 1-6\times 1}{6\times 8} \right)}^{-1}} \\
 & \Rightarrow {{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}={{\left( \dfrac{1}{6} \right)}^{-1}}+{{\left( \dfrac{2}{48} \right)}^{-1}} \\
 & \Rightarrow {{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}=={{\left( \dfrac{1}{6} \right)}^{-1}}+{{\left( \dfrac{1}{24} \right)}^{-1}} \\
\end{align}$
Again, using the exponential formula ${{a}^{-1}}=\dfrac{1}{a}$ in the above equation, then we will have
$\begin{align}
  & {{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}=\dfrac{1}{\dfrac{1}{6}}+\dfrac{1}{\dfrac{1}{24}} \\
 & \Rightarrow {{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}=\dfrac{6}{1}+\dfrac{24}{1} \\
 & \Rightarrow {{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}=30 \\
\end{align}$
Hence the simplified value of the expression ${{\left( {{2}^{-1}}-{{3}^{-1}} \right)}^{-1}}+{{\left( {{6}^{-1}}-{{8}^{-1}} \right)}^{-1}}$ is $30$ .

Note: For the first problem we have only used two of exponential formulas which are ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$ and ${{a}^{0}}=1$ to solve the problem. There are more exponential formulas which will be helpful while solving this type of problem. Some of the most useful formulas are given below.
${{a}^{m}}\times {{b}^{m}}={{\left( ab \right)}^{m}}$ , ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ .
For the second problem we have all the powers of the values as $-1$ only. So we have used the exponential formula ${{a}^{-1}}=\dfrac{1}{a}$ and simplified the equation by using basic mathematical operations. If we have the different powers for the values, then we need to use different exponential formulas and simplify the equation by using basic mathematical operations. Some of the exponential formulas are given below.
${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ , ${{a}^{1}}=a$ .