Answer
99/25
Work Step by Step
Exponents are used to represent repeated multiplication. For example, if a number a is raised to the power n, where n is an integer, it implies the number a is multiplied to itself n times.
The above statement can be shown mathematically as follows, assuming a to be equal to 2 and n to be equal to 5:
\[{{2}^{5}}=2\cdot 2\cdot 2\cdot 2\cdot 2\]
where the dot \[\left( . \right)\] represents multiplication.
Now, apply the property mentioned above to solve the given problem:
\[\begin{align}
& \frac{{{2}^{6}}}{{{2}^{4}}}-\frac{{{5}^{4}}}{{{5}^{6}}}=\frac{2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2}{2\cdot 2\cdot 2\cdot 2}-\frac{5\cdot 5\cdot 5\cdot 5}{5\cdot 5\cdot 5\cdot 5\cdot 5\cdot 5} \\
& =\frac{\cdot \cdot \cdot \cdot 2\cdot 2}{\cdot \cdot \cdot }-\frac{\cdot \cdot \cdot }{\cdot \cdot \cdot \cdot 5\cdot 5} \\
& =\frac{2\cdot 2}{1}-\frac{1}{5\cdot 5} \\
& =\frac{4}{1}-\frac{1}{25}
\end{align}\]
If a number is expressed in the form of \[\frac{p}{q}\], where p and q are integers, and \[q\ne 0\], then that number is said to be a fraction or a rational number.
The rational numbers \[\frac{4}{1}\] and \[\frac{1}{25}\] obtained above have denominators 1 and 25, respectively, which are different.
So, to add the above rational numbers, they should first be modified so that their denominators become same because, then, they can be easily added using the following property:
So, to modify the obtained rational numbers so that their denominators become same, first find the smallest number that is the multiple of the denominators of both the obtained rational numbers.
The smallest number that is the multiple of both 1 and 25 is 25. So, the denominators of the obtained rational numbers should somehow be multiplied with some numbers so that they become 25 as well as the effect of obtained rational numbers, on the original difference\[\frac{4}{1}-\frac{1}{25}\], still remains same.
To do that, follow the below procedure:
For any non-zero number d,\[\frac{d}{d}\] is always 1. And multiplying any rational number \[\frac{a}{b}\] with 1 or, equivalently, \[\frac{d}{d}\] will always result in the same rational number \[\frac{a}{b}\]. This means multiplying any rational number \[\frac{a}{b}\] with \[\frac{d}{d}\] will not affect the rational number itself.
So, the rational number \[\frac{4}{1}\] can be multiplied by \[\frac{25}{25}\] so that its denominator becomes25 as well as its effect on the original difference still remains same.
Hence, \[\frac{4}{1}=\frac{4}{1}\cdot \frac{25}{25}=\frac{4\cdot 25}{1\cdot 25}=\frac{100}{25}\]
And the denominator of the second rational number, \[\frac{1}{25}\], is already 25. So, it need not be modified.
Hence,
\[\begin{align}
& \frac{4}{1}-\frac{1}{25}=\frac{100}{25}-\frac{1}{25} \\
& =\frac{100-1}{25} \\
& =\frac{99}{25}
\end{align}\]
The above rational number is already reduced to its lowest terms because 99 and 25 do not have any common divisors except 1.
Hence,
\[\begin{align}
& \frac{{{2}^{6}}}{{{2}^{4}}}-\frac{{{5}^{4}}}{{{5}^{6}}}=\frac{4}{1}-\frac{1}{25} \\
& =\frac{99}{25}
\end{align}\]
