I'm teaching myself basic algebra from a book and am stuck on a question. In the current section it is about expressing numbers as powers of the same base. So $9$ maybe expressed as $3^2$.
Another e.g. is $${4^x \over 2^x} = {{2^2}^x \over 2^x}$$ and then I could use the indicie rules.
So the question I am stuck on is $${6^{2n}+1 \over 4^{n+2} \cdot 3^{2n-1}}$$
I have tried several different numbers but can find a common base that is shared by $6$, $4$ and $3$.
Is there a basic rule of some kind that I am missing ? Or is there something else at play here ? The book I am working from is a book for high school student. Regrettably my years at high school were not very academic ones so I'm trying to rectify this.
$\endgroup$1 Answer
$\begingroup$Since $6=2\times 3$ and $4=2^2$, you can use $2$ and $3$ as the bases.
$$\frac{6^{2n+1}}{4^{n+2}\cdot 3^{2n-1}}=\frac{2^{2n+1}\cdot 3^{2n+1}}{2^{2(n+2)}\cdot 3^{2n-1}}=2^{(2n+1)-(2n+4)}\cdot 3^{(2n+1)-(2n-1)}=2^{-3}\cdot 3^2=\frac{9}{8}.$$
$\endgroup$ 2
