To understand subtraction of exponents , let's start from a simple example.
Example . Suppose, we want to find 2 7 2 4 \frac{{{{2}}^{{7}}}}{{{{2}}^{{4}}}} 2 4 2 7 .
We already learned about positive integer exponets , so we can write, that 2 7 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 {{2}}^{{7}}={2}\cdot{2}\cdot{2}\cdot{2}\cdot{2}\cdot{2}\cdot{2} 2 7 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 and 2 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 {{2}}^{{4}}={2}\cdot{2}\cdot{2}\cdot{2} 2 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 .
So, 2 7 2 4 = 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 2 ⋅ 2 = 2 3 \frac{{{{2}}^{{7}}}}{{{{2}}^{{4}}}}=\frac{{{2}\cdot{2}\cdot{2}\cdot{\color{red}{{{2}\cdot{2}\cdot{2}\cdot{2}}}}}}{{{\color{red}{{{2}\cdot{2}\cdot{2}\cdot{2}}}}}}={2}\cdot{2}\cdot{2}={{2}}^{{3}} 2 4 2 7 = 2 ⋅ 2 ⋅ 2 ⋅ 2 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 ⋅ 2 = 2 ⋅ 2 ⋅ 2 = 2 3 .
Let's see what have we done. We counted number of 2's in 2 7 {{2}}^{{7}} 2 7 , then counted number of 2's in 2 4 {{2}}^{{4}} 2 4 . Since we divided, we canceled common terms. Note, that 7 − 4 = 3 {7}-{4}={3} 7 − 4 = 3 .
It appears, that this rule works not only for positive integer exponents, it works for any exponent.
Rule for subtracting exponents : a m a n = a m − n \color{purple}{\frac{a^m}{a^n}=a^{m-n}} a n a m = a m − n .
Word of caution. It doesn't work, when bases are not equal.
For example, 4 5 3 2 = 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4 3 ⋅ 3 \frac{{{{4}}^{{5}}}}{{{{3}}^{{2}}}}=\frac{{{4}\cdot{4}\cdot{4}\cdot{4}\cdot{4}}}{{{3}\cdot{3}}} 3 2 4 5 = 3 ⋅ 3 4 ⋅ 4 ⋅ 4 ⋅ 4 ⋅ 4 which is neither 4 3 {{4}}^{{3}} 4 3 nor 3 3 {{3}}^{{3}} 3 3 .
Word of caution. Above rule doesn't work for addition and subtraction.
For example, 2 7 − 2 4 ≠ 2 3 {{2}}^{{7}}-{{2}}^{{4}}\ne{{2}}^{{3}} 2 7 − 2 4 = 2 3 , because 2 7 − 2 4 = 128 − 16 = 112 {{2}}^{{7}}-{{2}}^{{4}}={128}-{16}={112} 2 7 − 2 4 = 128 − 16 = 112 and 2 3 = 8 {{2}}^{{3}}={8} 2 3 = 8 . Clearly, 112 ≠ 8 {112}\ne{8} 112 = 8 .
Let's go through a couple of examples.
Example 2 . Find 2 3 2 − 5 \frac{{{{2}}^{{3}}}}{{{{2}}^{{-{5}}}}} 2 − 5 2 3 .
It doesn't matter, that exponent is negative .
Just proceed as always: 2 3 2 − 5 = ( 2 3 − ( − 5 ) ) = 2 3 + 5 = 2 8 \frac{{{{2}}^{{3}}}}{{{{2}}^{{-{5}}}}}={\left({{2}}^{{{3}-{\left(-{5}\right)}}}\right)}={{2}}^{{{3}+{5}}}={{2}}^{{8}} 2 − 5 2 3 = ( 2 3 − ( − 5 ) ) = 2 3 + 5 = 2 8 .
Even when exponents are fractional , we use the same rule!
Example 3 . Find 3 1 4 3 2 3 \frac{{{{3}}^{{\frac{{1}}{{4}}}}}}{{{{3}}^{{\frac{{2}}{{3}}}}}} 3 3 2 3 4 1 .
3 1 4 3 2 3 = 3 1 4 − 2 3 = 3 − 5 12 = 1 3 5 12 \frac{{{{3}}^{{\frac{{1}}{{4}}}}}}{{{{3}}^{{\frac{{2}}{{3}}}}}}={{3}}^{{\frac{{1}}{{4}}-\frac{{2}}{{3}}}}={{3}}^{{-\frac{{5}}{{12}}}}=\frac{{1}}{{{3}}^{{\frac{{5}}{{12}}}}} 3 3 2 3 4 1 = 3 4 1 − 3 2 = 3 − 12 5 = 3 12 5 1 .
We can handle radicals, also, because radicals can be rewritten with the help of exponent.
Example 4 . Rewrite, using positive exponent: 3 8 1 3 2 7 \frac{{\sqrt[{{8}}]{{{3}}}}}{{\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}} 7 3 2 1 8 3 .
First we rewrite numbers, using exponents and then apply the rule:
3 8 1 3 2 7 = 3 1 8 3 − 2 7 = 3 1 8 3 − 2 7 = 3 1 8 − ( − 2 7 ) = 3 23 56 \frac{{\sqrt[{{8}}]{{{3}}}}}{{\sqrt[{{7}}]{{\frac{{1}}{{{3}}^{{2}}}}}}}=\frac{{{{3}}^{{\frac{{1}}{{8}}}}}}{{\sqrt[{{7}}]{{{{3}}^{{-{2}}}}}}}=\frac{{{{3}}^{{\frac{{1}}{{8}}}}}}{{{{3}}^{{-\frac{{2}}{{7}}}}}}={{3}}^{{\frac{{1}}{{8}}-{\left(-\frac{{2}}{{7}}\right)}}}={{3}}^{{\frac{{23}}{{56}}}} 7 3 2 1 8 3 = 7 3 − 2 3 8 1 = 3 − 7 2 3 8 1 = 3 8 1 − ( − 7 2 ) = 3 56 23 .
Finally, we can see now, why a 0 = 1 {{a}}^{{0}}={1} a 0 = 1 .
Indeed, a 0 = a n − n = a n a n = 1 {{a}}^{{0}}={{a}}^{{{n}-{n}}}=\frac{{{{a}}^{{n}}}}{{{{a}}^{{n}}}}={1} a 0 = a n − n = a n a n = 1 .
Now, it is time to exercise.
Exercise 1 . Find 3 5 3 2 \frac{{{{3}}^{{5}}}}{{{{3}}^{{2}}}} 3 2 3 5 .
Answer : 3 3 = 27 {{3}}^{{3}}={27} 3 3 = 27 .
Exercise 2 . Can we use rule for adding exponents to find 5 5 3 5 \frac{{{{5}}^{{5}}}}{{{{3}}^{{5}}}} 3 5 5 5 ?
Answer : No, bases are not equal.
Exercise 3 . Find 4 5 3 4 2 3 \frac{{{{4}}^{{\frac{{5}}{{3}}}}}}{{{{4}}^{{\frac{{2}}{{3}}}}}} 4 3 2 4 3 5 .
Answer : 4 {4} 4 .
Exercise 4 . Find 3 2 3 − 1 5 \frac{{{{3}}^{{2}}}}{{{{3}}^{{-\frac{{1}}{{5}}}}}} 3 − 5 1 3 2 .
Answer : 3 2 + 1 5 = 3 11 5 = 3 11 5 {{3}}^{{{2}+\frac{{1}}{{5}}}}={{3}}^{{\frac{{11}}{{5}}}}={\sqrt[{{5}}]{{{{3}}^{{11}}}}} 3 2 + 5 1 = 3 5 11 = 5 3 11 .
Exercise 5 . Find 1 27 7 9 8 \frac{{\sqrt[{{7}}]{{\frac{{1}}{{27}}}}}}{{\sqrt[{{8}}]{{{9}}}}} 8 9 7 27 1 .
Answer : 3 − 3 7 3 2 8 = 1 3 19 28 \frac{{\sqrt[{{7}}]{{{{3}}^{{-{3}}}}}}}{{\sqrt[{{8}}]{{{{3}}^{{2}}}}}}=\frac{{1}}{{{3}}^{{\frac{{19}}{{28}}}}} 8 3 2 7 3 − 3 = 3 28 19 1 .