2^2x-6(2^x)+8=0

asked by guest
on Nov 18, 2024 at 10:46 am



You asked:

Solve the equation 22x−6⋅2x+8=0{2}^{2} x - 6 \cdot {2}^{x} + 8 = 0 for the variable xx.

MathBot Answer:

The 2 solutions to the equation are: x=0.380148278929398106201375688025455377239737583697−2log⁡(2)log⁡(2)≈−1.451562x=2.07944154167983592825169636437452964934702847231−2log⁡(2)log⁡(2)≈1.0\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}

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