2^2x-6(2^x)+8=0
You asked:
Solve the equation \({2}^{2} \cdot x - 6 \cdot {2}^{x} + 8 = 0\) for the variable \(x\).
MathBot Answer:
The 2 solutions to the equation are: \[\begin{aligned}x &= \frac{0.380148278929398106201375688025455377239737583697 - 2 \log{\left(2 \right)}}{\log{\left(2 \right)}} \approx -1.451562\\x &= \frac{2.07944154167983592825169636437452964934702847231 - 2 \log{\left(2 \right)}}{\log{\left(2 \right)}} \approx 1.0\end{aligned}\]