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.
$\endgroup$ 03 Answers
$\begingroup$HINT:
First of all, notice that $15^{x-1} \equiv \frac{1}{15} \times 15^x$, and hence: $$3^{2x} - 34\left(15^{x-1}\right)+5^{2x} \equiv 3^{2x} - \tfrac{34}{15}\left(15^x\right) + 5^{2x}$$
Second, notice that
$$u^2 - \tfrac{34}{15}uv + v^2 \equiv \tfrac{1}{15}(5u-3v)(3u-5v)$$
If you put $u=3^x$ and $v=5^x$, then notice that $u^2 = 3^{2x}$, $uv = 3^x\times 5^x = 15^x$ and $v^2 = 5^{2x}$.
Finally, you might find logarithms help you out.
$\endgroup$ $\begingroup$Divide by $3^{2x}$, and you have a quadratic equation in $(\frac 53)^x$.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$
