2^(2x+1) - 5 * 2^(2x) + 2 = 0
You asked:
Solve the equation \({2}^{2 x + 1} - 5 \cdot {2}^{2 x} + 2 = 0\) for the variable \(x\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}x &= \log{\left(\left(\frac{2}{3}\right)^{\frac{1}{2 \log{\left(2 \right)}}} \right)} \approx -0.29248125\\x &= \frac{\frac{\log{\left(\frac{2}{3} \right)}}{2} + i \pi}{\log{\left(2 \right)}} \approx -0.29248125 + 4.5323601 i\end{aligned}\]
\(i\) is the imaginary unit, defined as \(i^2 = -1\).