Question: > 
> Solve $$ 3-\frac{4}{9^x}-\frac{4}{81^x}=0 $$
> 
> 
> 

I had this question for an exam today and I want to find out if my answer was correct.

Answer: 
Substitute $t = 2 / 9^x$, then you get simple quadratic: $t^2 + 2t - 3 = 0$ ...

**UPDATE:** I want to update my answer regarding the solution of the equation:
The question does not specify if only real solutions are required. If that is the case there is a single solution: $ \ln2 / \ln9 $. However, this is not the only solution as the quadratic has one more root (-3) which will result in complex (infinitely many) solutions, i.e.:
$$ 9^x = - \frac 2 3 $$ or 
$$ x = \frac{ln (-1) + ln(2/3)}{ln9} $$

$ ln(-1)=(2k+1)\pi i$ for $ k \in \mathbb{Z} $, which give us infinitely many solutions to the equation:

$$ x=\frac{ln(2/3)}{ln9} + i \frac{(2k+1) \pi}{ln9} $$
So if we let the common denominator be $81^x$, we would get this:

$$\frac{81^x\times 3 - 4\times9^x - 4}{81^x} = 0$$

Use the quadratic formula on the numerator while letting $y = 9^x$, and we get

$$y = 9^x = \frac{2\pm 4}{3}$$

So $$x = \frac{\ln 2}{\ln 9}$$ 

since $\ln \frac{-2}{3}$ does not exist.