Solve-the-equation-2-x-8x-4-

Question Number 100879 by Tony6400 last updated on 29/Jun/20

Solve the equation 2^{x} + 8 x = 4

Commented by bemath last updated on 29/Jun/20

approximation x = 0.342

Answered by 1549442205 last updated on 07/Jul/20

\Leftrightarrow 2^{x - 2} + 2 x - 1 = 0 . Putting f (x) = 2^{x - 2} + 2 x - 1

f^{'} (x) = 2^{x - 2} \ln 2 + 2 > 0 \forall x \in (- \infty; + \infty)

f (x) = 0 \Leftrightarrow x = 0.3416043768

Since f (x) is increasing, the given equation has

only root x = 0.3416043768

Answered by mr W last updated on 29/Jun/20

2^{x} = 8 (\frac{1}{2} - x)

e^{x \ln 2} = 8 (\frac{1}{2} - x)

\frac{\sqrt{2} \ln 2}{8} = [(\frac{1}{2} - x) \ln 2] e^{(\frac{1}{2} - x) \ln 2}

(\frac{1}{2} - x) \ln 2 = W (\frac{\sqrt{2} \ln 2}{8})

\Rightarrow x = \frac{1}{2} - \frac{1}{\ln 2} W (\frac{\sqrt{2} \ln 2}{8})

\approx \frac{1}{2} - \frac{0.109791}{\ln 2} = 0.341605

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