$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n-2}}{6^{n+1}} = -\frac{5}{27}$

Question

$$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n-2}}{6^{n+1}} = -\frac{5}{27}$$

I currently don't have any idea about where to start. Can you take a look?

My attempt:

$$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n}\cdot 3}{6^{n+1}} = -\frac{5}{27}$$

$$\frac{3^n (-5 \cdot + 3^{-2})}{6^{n+1}} = -\frac{5}{27}$$

Regards,

Answer 1

$$\frac{3^n - 5 \cdot 3^{n-1} + 3^{n-2}}{6^{n+1}} = -\frac{5}{27}$$

$$\frac{3^n (1- \frac{5}3 + \frac19)}{6^{n}\cdot 6} = -\frac{5}{27}$$

$$\left(\frac12\right)^n\frac{ (1- \frac{5}3 + \frac19)}{ 6} = -\frac{5}{27}$$

Try to isolate $\left(\frac12\right)^n$ and then take logarithm to complete the question.

Answer 2

Use the substitution $K = 3^{n-2}$ and exponent/power rules.