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

1 Answer
Cesareo R.
May 13, 2016

The equation has not real solution.

Explanation:

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

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