Find the common ratio given $n$ via sum

Question

Let $S_n$ be the sum of the first $n$ elements of the geometric sequence. We know that $$\log_3\left(\frac{S_n}{2}+1\right)=n$$ Given this, determine the common ratio $r$.

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My attempt

We've got $$\log_3\left(\frac{S_n}{2}+1\right)=n$$ We can remove the logarithm by raising both sides to $3$. $$\frac{S_n}{2}+1=3^n$$ Let's subtract $1$ from both sides $$\frac{S_n}{2}=3^n-1$$ Now, let's multiply everything by $2$ $$S_n=2 \cdot 3^n-2$$ Now, let's rewrite $S_n$ $$a_1\left(\frac{1-r^n}{1-r}\right)=2\cdot3^n-2$$ I can continue, but it doesn't lead me anywhere.

Answer 1

$S_1=2(3^1-1) $ and $S_2=2(3^2-1)$

$\Rightarrow a_1=4 \ and \ a_2=S_2-S_1= 16-4=12 \Rightarrow r=\frac{a_2}{a_1}=3$

Answer 2

The formula is valid for any $n\ge 1$. Plugging in $n=1$ yields $a_1=4$. For $n=2$, you get $r=3$. If you use these values, $$4\left(\frac{1-3^n}{1-3}\right)=4\frac{3^n-1}{2}=2\cdot3^n-2$$ This verifies your last identity.