# Algebraic Rule Explanation

Can someone please explain to me why $\dfrac{3^{n-1}}{3^n}$ is equal to $\dfrac{1}{3}$?

Also are there any videos or resources that you can provide to teach me some of these rules?

Thanks you.

• In general $\frac{a^n}{a^m}=a^{n-m}$. – Xam Nov 23 '16 at 17:44
• I've written a prototype for a web page that you may find helpful. mathontrack.comze.com/exponentials2.html – John Joy Nov 24 '16 at 13:49

$$\frac{3\cdot3\cdot3\cdot3\cdot3\cdot3}{3\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3}=\frac13,$$ no ?

• @amWhy: the case $n=7$ is quite sufficient to allow anyone to generalize as the pattern is obvious. This is a kind of answer "without words", IMO more efficient than other algebraic manipulations. – Yves Daoust Nov 23 '16 at 19:49
• @amWhy: cheers ! – Yves Daoust Dec 11 '16 at 18:08

we have $$\frac{3^{n-1}}{3^n}=3^{n-1-n}=3^{-1}=\frac{1}{3}$$

Notice that

$$\frac{3^{n-1}}{3^n}=\frac{3^{n-1}}{3 \cdot 3^{n-1}} \\ =\frac{1}{3}.$$ The same technique can be used to prove the general formula

$$\frac{a^n}{a^m}=a^{n-m}.$$

In general $$x^b / x^c = x^{b - c}$$

now take $x=3$,$b = n-1$, and $c = n$ and you will get

$$\frac{3^{n-1}}{3^n} = 3^{(n -1) - n} = 3^{-1} = \frac{1}{3}$$