How do you solve $4 x - 2^{x + 1} - 15 = 0$ $4x - 2^(x+1) - 15 =0$ ?

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How do you solve $4 x - 2^{x + 1} - 15 = 0$ $4x - 2^(x+1) - 15 =0$ ?

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Answer 1 · 2016-05-13T21:00:32

The equation has not real solution.

Explanation:

Consider the equation $f = 4 x - 2^{x + 1}$ $f = 4x-2^(x+1)$ . This equation has a maximum for $x = 1.5287$ $x = 1.5287$ This poin is the critical point obtained solving $\frac{d f}{d x} = 4 - 2^{1 + x} ln 2 = 0$ $(df)/dx = 4 - 2^(1 + x) Ln 2 = 0$ giving $x = \frac{ln (\frac{4}{ln (2)})}{ln (2)} - 1 = 1.52877$ $x = Ln(4/Ln(2))/Ln(2)-1 = 1.52877$ . Substituting this value in $f (1.52877) = 0.344285$ $f(1.52877) = 0.344285$ . This value is lower than $15$ $15$ so no real solution is expected.

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How do you solve $4 x - 2^{x + 1} - 15 = 0$ $4x - 2^(x+1) - 15 =0$ ?

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