# How to prove that $\frac{a^n}{a^m}$ is equal to $a^{n-m}$? [closed]

How to prove that $\dfrac{a^n}{a^m}$ is equal to $a^{n-m}$?

Assuming $n>m>0$. $$\require{cancel} \frac{a^n}{a^m}=\frac{\overbrace{a\cdot a\cdot a\cdot a\cdot a\cdots a}^{\large n\text{ times}}}{\underbrace{a\cdot a\cdot a\cdots a}_{\large m\text{ times}}}=\frac{\color{red}{\cancel{\color{black}{a\cdot a\cdot a\cdots a}}}\overbrace{\cdot a\cdot a\cdots a}^{\large n-m\text{ times}}}{\color{red}{\cancel{\color{black}{a\cdot a\cdot a\cdots a}}}}=a^{n-m}.$$

• Thank you. It is exactly what I have been looking for. Commented Jun 10, 2014 at 14:06
• @cherry8.8vanilla You're welcome. Commented Jun 10, 2014 at 14:07

Let $n>m$, then $n$ can be written as $m+r$

$\therefore \Large\frac{a^n}{a^m}=\frac{a^{m+r}}{a^m}=\frac{a^m*a^r}{a^m}=a^r$

but, $n=m+r$ $\Longrightarrow$ $r=n-m$

$\therefore \Large\frac{a^n}{a^m}=a^{n-m}$

depending what you area allowed to use...$\frac{a^n}{a^m} = a^n \cdot \frac{1}{a^m} = a^n \cdot a^{-m}= a^{n-m}$.

EDIT once again, this depends on what you are allowed to use. This is true for $a>0$. Log both sides. On RHS you get $(n-m) \log a$, on LHS $n \log a - m \log a = (n-m) \log a$. Now use the $a \log x =\log x^a$ property of the log function and exponentiate back.

• I need to show how to get from $\frac{a^n}{a^m}$ or $a^n \cdot \frac{1}{a^m}$ or $a^n \cdot a^{-m}$ this $a^{n-m}$. Commented Jun 10, 2014 at 13:00
• pls see the edit
– Alex
Commented Jun 10, 2014 at 13:02