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I am having a ridiculous amount of trouble solving questions of this style - this one in particular, I know the answer is $6^{2n-4}$ but I can't get there!!!!

$ \frac {36^{2n}.6^{n+2}} {216^{n+2}}$

Could someone please help me out with some steps involved? I would be extremely grateful!

I understand the rules $a^x b^x = (ab)^x $ and $x^a x^b = x^{a+b} $ I just don't understand why I'm not able to get this.

These are my steps I have been taking:

$\frac {3^{2n}.12^{2n}.2^{n+2}.3^{n+2}} {12^{n+2}.18^{n+2}}$

$ \frac {3^{2n}.3^{2n}.4^{2n}.2^{n+2}.3^{n+2}} {4^{n+2}.3^{n+2}.3^{n+2}.3^{n+2}.2^{n+2}} $

I've simplified in this manner and then have tried cancelling and am coming out with completely wrong answers every time.

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    $\begingroup$ Since $36=6^2$ and $216=6^3$, you don't really need to factor. $\endgroup$ – André Nicolas Aug 4 '14 at 21:56
  • $\begingroup$ André is right. You do not need to factor 6(=2x3). $\endgroup$ – mike Aug 4 '14 at 21:59
  • $\begingroup$ That makes sense.Thanks for pointing that out!! :) $\endgroup$ – Dani Aug 4 '14 at 23:18
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Since $36=6^2$ and $216=6^3$, you don't really need to factor.

However, we will do it in factoring style. Although it is in this case inefficient, in other situations it is the right thing to do.

For problems of this general kind, you should express things as a product of powers of distinct primes. (In your calculation, both powers of $2$ and powers of $4$ remained. Since $4=2^2$, these should have been combined.)

On top, we have $36^{2n}$. This is $(2^2\cdot 3^2)^{2n}$, which is $2^{4n}\cdot 3^{4n}$. The term $6^{n+2}$ is $2^{n+2}\cdot 3^{n+2}$. So the top is $2^{5n+2}\cdot 3^{5n+2}$.

Since $216=2^3\cdot 3^3$, the bottom is $2^{3n+6}\cdot 3^{3n+6}$.

Divide. We get $2^{2n-4}\cdot 3^{2n-4}$, which can be rewritten as $6^{2n-4}$.

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  • $\begingroup$ Thank you! Makes perfect sense when explained like that. $\endgroup$ – Dani Aug 4 '14 at 23:19
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$$ \frac {36^{2n}.6^{n+2}} {216^{n+2}}=\frac {(6^2)^{2n}.6^{n+2}} {(6^3)^{n+2}}=\frac{6^{2\cdot2n+(n+2)}}{6^{3\cdot(n+2)}}=\frac{6^{5n+2}}{6^{3n+6}}=6^{(5n+2)-(3n+6)}=6^{2n-4}$$

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  • $\begingroup$ Excellent this is exactly what I need to see to help me understand. Thank you very much! $\endgroup$ – Dani Aug 4 '14 at 23:20
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$$\frac {36^{2n}\cdot6^{n+2}} {216^{n+2}} \equiv \frac{(6^2)^{2n} \cdot 6^{n+2}}{(6^3)^{n+2}} \equiv \frac{6^{2(2n)} \cdot 6^{n+2}}{6^{3(n+2)}} \equiv \frac{6^{4n} \cdot 6^{n+2}}{6^{3n+6}} $$ $$\equiv \frac{6^{4n+n+2}}{6^{3n+6}} \equiv \frac{6^{5n+2}}{6^{3n+6}} \equiv 6^{(5n+2)-(3n+6)} \equiv \color{green}{6^{2n-4}} \equiv 6^{2(n-2)} \equiv (6^2)^{n-2}$$

$$\equiv \;\bbox[3pt,color:black;border-radius:4px;box-shadow:2px 2px 4px green]{36^{n-2}}\;$$

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