Question

How do you simplify $\dfrac{7a}{\sqrt[5]{4{{a}^{7}}{{b}^{13}}}}$?

Answer 1

Hint: The given expression in the above question contains a fraction whose numerator is already in the simplified form but the denominator is complex. For simplifying the expression, we first need to simplify the denominator. We can write the denominator as \[{{\left( 4{{a}^{7}}{{b}^{13}} \right)}^{\dfrac{1}{5}}}\]. Then, we have to apply the property of the exponent given by ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$ so that the fractional power $\dfrac{1}{5}$ comes on each of the three terms $4$, ${{a}^{7}}$ and ${{b}^{13}}$. Then, we need to take these on the numerator by using the negative exponent property $\dfrac{1}{{{a}^{m}}}={{a}^{-m}}$. Finally, using the property of the common bases given by ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$, we will get a simplified expression.

Complete step by step solution:
Let us write the expression given in the above question as
\[\Rightarrow E=\dfrac{7a}{\sqrt[5]{4{{a}^{7}}{{b}^{13}}}}........\left( i \right)\]
Considering the denominator of the above expression, we have
$\Rightarrow D=\sqrt[5]{4{{a}^{7}}{{b}^{13}}}$
The denominator can also be written as
$\Rightarrow D={{\left( 4{{a}^{7}}{{b}^{13}} \right)}^{\dfrac{1}{5}}}$
Now, we know that ${{\left( ab \right)}^{m}}={{a}^{m}}{{b}^{m}}$. Applying this property above, we can rewrite the above expression as
$\Rightarrow D={{4}^{\dfrac{1}{5}}}{{a}^{\dfrac{7}{5}}}{{b}^{\dfrac{13}{5}}}$
Substituting this in (i) we get
\[\Rightarrow E=\dfrac{7a}{{{4}^{\dfrac{1}{5}}}{{a}^{\dfrac{7}{5}}}{{b}^{\dfrac{13}{5}}}}\]
Now, using the property of the negative exponent, given by $\dfrac{1}{{{a}^{m}}}={{a}^{-m}}$ we can rewrite the above expression as
\[\begin{align}
  & \Rightarrow E=\dfrac{7a}{{{4}^{\dfrac{1}{5}}}}\left( {{a}^{-\dfrac{7}{5}}}{{b}^{-\dfrac{13}{5}}} \right) \\
 & \Rightarrow E=\dfrac{7}{{{4}^{\dfrac{1}{5}}}}\times a\times {{a}^{-\dfrac{7}{5}}}\times {{b}^{-\dfrac{13}{5}}} \\
 & \Rightarrow E=\dfrac{7{{b}^{-\dfrac{13}{5}}}}{{{4}^{\dfrac{1}{5}}}}\times a\times {{a}^{-\dfrac{7}{5}}} \\
\end{align}\]
Now, using the property of the common base, given by ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$, we can write the above expression as
\[\begin{align}
  & \Rightarrow E=\dfrac{7{{b}^{-\dfrac{13}{5}}}}{{{4}^{\dfrac{1}{5}}}}\times {{a}^{1-\dfrac{7}{5}}} \\
 & \Rightarrow E=\dfrac{7{{b}^{-\dfrac{13}{5}}}}{{{4}^{\dfrac{1}{5}}}}\times {{a}^{\dfrac{5-7}{5}}} \\
 & \Rightarrow E=\dfrac{7{{b}^{-\dfrac{13}{5}}}}{{{4}^{\dfrac{1}{5}}}}\times {{a}^{\dfrac{-2}{5}}} \\
 & \Rightarrow E=\dfrac{7{{a}^{\dfrac{-2}{5}}}{{b}^{-\dfrac{13}{5}}}}{{{4}^{\dfrac{1}{5}}}} \\
\end{align}\]
Hence, the given expression is simplified.

Note: We can also simplify the given expression by taking $a$ from the numerator to the denominator using the same property of negative exponent, as used above. The main motive of simplification is to solve the exponent of the bases common to the numerator and the denominator.