The function is of the form
#f(x) = ax-2^x#
and so
#f^'(x) = a-(ln 2) 2^x #
Thus #f^'(b) = 0 implies#
#0 = a-(ln 2) 2^b#
#2^b = a/ln 2 implies b = log_2(a/ln 2) = ln(a/ln2)/ln 2#
To complete the answer we need to determine the value of #a# explicitly.
#a = root{3}{6+root{3}{6+root{3}{6+...}}} implies #
#a^3 = 6+root{3}{6+root{3}{6+root{3}{6+...}}} = 6+a implies#
#a^3-a-6 = 0 implies#
#a^3-2a^2 +2a^2-4a +3a-6 = 0 implies#
#(a^2+2a+3)(a-2) = 0#
Since #a^2+2a+3=0# can not have real roots, we must have #a = 2#
Thus
#b = ln(2/ln 2)/ln 2 = frac{ln 2 - ln(ln2)}{ln 2} implies#
#b = 1-ln(ln 2)/ln 2#
