I tried to solve this exponential equation but got only one of the two solutions right.
This is what I did: $$ 17 \cdot 64^{x} - 4 \cdot 8^{x} - 164 = - 42 \cdot 8^{1-x} + 2 \cdot 8^{2-x} $$ I set everything to one side: $$ \Leftrightarrow 17 \cdot 64^{x} - 4 \cdot 8^{x} - 164 + 42 \cdot 8^{1-x} - 2 \cdot 8^{2-x} = 0 \\ \Leftrightarrow 17 \cdot \left({8^x}\right)^2 - 4 \cdot 8^{x} - 164 + 42 \cdot 8^{1} \cdot 8^{x^{-1}} - 2 \cdot 8^{2} \cdot 8^{-x} = 0 $$ I then replaced $8^x$ with y $$ \Leftrightarrow 17 \cdot y^2 - 4y - 164 + 336 y^{-1} - 128 y^{-1} = 0 \\ \Leftrightarrow \frac{17 \cdot y³ - 4y² - 164y + 336 - 128}{y} = 0 \\ \Leftrightarrow 17y^3 - 4y^2 - 164y + 208 = 0 \ \text{ ($y$ is NOT 0)} \\ (\ldots) \\ \Leftrightarrow y = 2\text{ or }17y^2 + 30y - 104 = 0 \\ D = b^2 - 4ac\\ = 7972 \\ y = 1,743707056\ \text{ (also a negative solution which I leave behind)} \\ y = 2 \text{ or } y = 1,743707056 \\ \Leftrightarrow 8^x = 2 \text{ or } 8^x = 1,743707056 \\ \Leftrightarrow \log 8^x = 2 \text{ or } \log 8^x = 1,743707056 \\ \Leftrightarrow \ldots \\ \Leftrightarrow x = \frac13 \text{ or } x = 0,26739 $$ $x = \frac13$ is a correct solution $x = 0,26729$ is a false solution, but I can't see where I'm wrong if someone can help.
Thank you in advance.