Question

How do you evaluate and simplify $\dfrac{{{64}^{\dfrac{5}{9}}}\cdot {{64}^{\dfrac{2}{9}}}}{{{4}^{\dfrac{3}{4}}}}$ ?

Answer 1

Hint: To evaluate and solve these types of questions we will use some laws of mathematics. If the fractional exponents are in multiplied form then the first law is; ${{x}^{a}}\cdot {{x}^{b}}={{x}^{a+b}}$. In upper part of the equation let $x$ is equals to 64 and $a$ is equals to $\dfrac {5} {9} $ and $b$is equals to $\dfrac {2} {9} $. If the fractional exponents are in ratio form like$\dfrac{{{x}^{a}}}{{{x}^{b}}}$ then the second formula for solving is $\dfrac{{{x}^{c}}}{{{x}^{d}}}={{x}^{c-d}}$ and if the fractional exponents is as ${{\left( {{x}^{e}} \right)}^{f}}$ then the third formula for this is:${{\left( {{x}^{e}} \right)}^{f}}={{x}^{e\cdot f}}$ .

Complete step by step solution:
The given equation is
$\Rightarrow \dfrac{{{64}^{\dfrac{5}{9}}}\cdot {{64}^{\dfrac{2}{9}}}}{{{4}^{\dfrac{3}{4}}}}$
Now first solve the upper part of the equation by using first law ${{x} ^ {a}}\cdot {{x} ^ {b}} = {{x} ^ {a+b}} $
$\begin {align}
  & \Rightarrow {{64} ^ {\dfrac{5}{9}}}\cdot {{64} ^ {\dfrac{2}{9}}} \\
 & \Rightarrow {{64} ^ {\dfrac{5}{9}+\dfrac{2}{9}}} \\
 & \Rightarrow {{64} ^ {\dfrac{7}{9}}} \\
\end{align}$
Now we can write $64= {{4} ^ {3}} $ so by putting this value in above equation we get,
$\Rightarrow {{\left ({{4} ^ {3}} \right)}^{\dfrac{7}{9}}}$
 by using the third formula ${{\left ({{x} ^ {e}} \right)}^{f}}={{x}^{e\cdot f}} $ we can write it as:
$\Rightarrow {{4} ^ {\dfrac{21}{9}}}$
Again we can simplify it by writing $4= {{2} ^ {2}} $ .so by putting this in above equation we get,
$\Rightarrow {{\left ({{2} ^ {2}} \right)}^{\dfrac{21}{9}}}$
Again by using the third formula ${{\left ({{x} ^ {e}} \right)}^{f}}={{x}^{e\cdot f}} $ we can write the above as:
$\begin {align}
  & \Rightarrow {{2} ^ {\dfrac{2\cdot 21} {9}}} \\
 & \Rightarrow {{2} ^ {\dfrac{42}{9}}} \\
\end{align}$
Now put the equivalent value of upper part in the given equation $\dfrac{{{64}^{\dfrac{5}{9}}}\cdot {{64}^{\dfrac{2}{9}}}}{{{4}^{\dfrac{3}{4}}}}$we get,
$\Rightarrow \dfrac{{{64}^{\dfrac{5}{9}}}\cdot {{64}^{\dfrac{2}{9}}}}{{{4}^{\dfrac{3}{4}}}}=\dfrac{{{2}^{\dfrac{42}{9}}}}{{{4}^{\dfrac{3}{4}}}}$
Here we can write $4= {{2} ^ {2}} $ in the above equation, we get,
$\Rightarrow \dfrac{{{2}^{\dfrac{42}{9}}}}{{{2}^{\dfrac{6}{4}}}}$
Now we will apply the second the formula $\dfrac {{{x} ^{c}}} {{{x} ^ {d}}} = {{x} ^{c-d}} $ then we get,
$\begin {align}
  & \Rightarrow \dfrac{{{2}^{\dfrac{42}{9}}}}{{{2}^{\dfrac{6}{4}}}} \\
 & \Rightarrow {{2} ^ {\dfrac{42}{9}-\dfrac{6}{4}}}........\left (1 \right) \\
 & \\
\end{align}$
Now by taking the LCM of $\dfrac {42} {9}-\dfrac {6} {4} $ we get,
$\begin {align}
  & \Rightarrow \dfrac{168-54}{36} \\
 & \Rightarrow \dfrac{114}{36} \\
 & \Rightarrow \dfrac{19}{6} \\
\end{align}$
Now put this value of power in equation (1) we get,
$\Rightarrow {{2} ^ {\dfrac{19}{6}}}$
Hence by using the laws of algebra we have reached to the solution of $\dfrac{{{64}^{\dfrac{5}{9}}}\cdot {{64}^{\dfrac{2}{9}}}}{{{4}^{\dfrac{3}{4}}}}$is equal to ${{2}^{\dfrac{19}{6}}}$.

Note: Here we can go wrong by writing and applying the laws of the algebra. So we should be careful by writing the laws. Without applying laws the process may go lengthy and complicated. So it is necessary to first simplify the equation and then apply the laws.