How can I simplify $\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$?

Question

This problem got me thinking hard:
$$\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$$ Considering the fact that the common base is '$3$', I decided to do this:
$$\log_3\left(\frac{(n+1)(n-1)}{(n+2)(n)}\right)$$ $$\log_3\left(\frac{n^2-n+n+1}{n^2+2n}\right)$$ $$\log_3\left(\frac{n^2 + 1}{n^2 + 2n}\right)$$ $$\log_3\left(\frac{1}{2n}\right)$$ I am not 100% sure if have simplified the problem. I need some clarification.

Answer 1

You're kidding? A sledgehammer to crack a nut?

Factor out $3^{n-1}$ in the numerator and denominator to get eventually $\dfrac13$.

Answer 2

Factor out by $3^n $ to get

$$\frac{3^n (3+3^{-1})}{3^n (3^2+3^0)} $$

$$\frac {3+3^{-1}}{10}=\frac {1}{3} $$

Answer 3

Why are taking logarithm? No need of that. It's just simple Indices.

$\dfrac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n} =\dfrac{3^{n-1}\left(3^2+1\right)}{3^n\left(3^2+1\right)} =\dfrac{3^{n-1}}{3^n} =\dfrac{1}{3}$

Answer 4

$$\frac{3^{n+1}+3^{n-1}}{3^{n+2}+3^n}=\frac{3^{n-1}(3^2+1)}{3^{n-1}(3^3+3)}=\frac{10}{30}=\frac 13$$ This shows your simplification is incorrect for a number of reasons:

• $\log(3^{n+1}+3^{n-1})\neq \log(3^{n+1})\log(3^{n-1})$. The $\log$ rule works the other way: $\log(a)+\log(b)=\log(ab)$
• You cannot simplify $\frac{n^2+1}{n^2+2n}$ to $\frac{1}{2n}$: big no-no. You can simplify multiplied expressions, not $summed$ expressions. For instance $\frac{n(n+1)}{(n+3)(n+1)}=\frac{n}{n+3}$ is fine.