<s>[INST] $3^{2x} - 34(15^{x-1}) + 5^{2x} = 0$ I've never seen anything like this. Do somebody have a way to solve it?
I've tried the basci exponential functions techniques but it does not work. Even substitution does not work...
I'm really intersted in learning, this is not a homework.
 [/INST] Another way to look at this is to consider that $ \ a^{2x} = (a^x)^2 \ , $ from the properties of exponents.  Suppose you had a binomial square $ \ (3^x + 5^x)^2 \ $ then:  its expansion would be 
$$ \ (3^x)^2 \ + \ 2 \cdot 3^x \cdot 5^x \ + \ (5^x)^2 \ = \ 3^{2x} \ + \ 2 \cdot (3 \cdot 5)^x \ + \ 5^{2x} \ = \ 3^{2x} \ + \ 2 \cdot 15^x \ + \ 5^{2x} \ \ . $$
Now, our equation is $ \ 3^{2x} - 34(15^{x-1}) + 5^{2x} = 0 \ , $
which we can arrange as
$$ 3^{2x} \ - \ 34 \ (15^x \cdot 15^{-1}) \ + \ 5^{2x} \ = \ 0   \ \ \Rightarrow  \ \ 3^{2x} \ - \ \frac{34}{15}  \cdot 15^x  \ + \ 5^{2x} \ = \ 0 $$
$$ \Rightarrow  \ \ 3^{2x} \ + \ (\frac{30}{15} -  \frac{64}{15})  \cdot 15^x  \ + \ 5^{2x} \ = \ 0 \ \ \Rightarrow \ \  (3^{2x} \ + \ 2   \cdot 15^x  \ + \ 5^{2x}) \ = \ \frac{64}{15}  \cdot 15^x $$
$$ \Rightarrow \ \  (3^x   \ + \ 5^x)^2  \  = \  \frac{64}{15}  \cdot \ 3^x \cdot \ 5^x \ = \ \frac{64}{15}  \cdot \ [ \ (\sqrt{3})^x \ ]^2 \cdot \ [ \ (\sqrt{5})^x\ ]^2 $$
$$  \Rightarrow \ \ \left( \ \left[ \ \sqrt{  \frac{3}{5}} \  \right]^x  + \  \left[ \ \sqrt{\frac{5}{3}} \  \right]^x  \ \right)^2  = \ \frac{64}{15} $$
$$\Rightarrow \ \  \left( \sqrt{\frac{3}{5}} \  \right)^x + \  \left( \ \sqrt{  \frac{5}{3}} \ \right)^x  \ = \ \frac{8}{\sqrt{15}}  \ \ = \ \frac{3 + 5}{\sqrt{3 \ \cdot \ 5}}  \ \ . $$
It's  reasonably clear from this result that $ \ x = 1 \ $ is a solution by simply adding the square-root ratios.  But because those ratios are reciprocals of one another, we see that $ \ x = -1 \ $ is also a solution.  So we have located the two roots of this "quadratic" equation.
This approach has little to recommend itself as a general technique: it simply works out nicely because of the "middle" coefficient in the equation.
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