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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.

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    $\begingroup$ In general $\frac{a^n}{a^m}=a^{n-m}$. $\endgroup$ – Xam Nov 23 '16 at 17:44
  • $\begingroup$ I've written a prototype for a web page that you may find helpful. mathontrack.comze.com/exponentials2.html $\endgroup$ – John Joy Nov 24 '16 at 13:49
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$$\frac{3\cdot3\cdot3\cdot3\cdot3\cdot3}{3\cdot3\cdot3\cdot3\cdot3\cdot3\cdot3}=\frac13,$$ no ?

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    $\begingroup$ @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. $\endgroup$ – Yves Daoust Nov 23 '16 at 19:49
  • $\begingroup$ @amWhy: cheers ! $\endgroup$ – Yves Daoust Dec 11 '16 at 18:08
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we have $$\frac{3^{n-1}}{3^n}=3^{n-1-n}=3^{-1}=\frac{1}{3}$$

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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

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

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

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

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

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