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

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