Find the solutions #x > 0 in RR# for #2^x + 2^(1+1/sqrt(x))=6#?

Despite that the solution is obviously 1, by inspection. an iterative

method with starter 1 disclosed this fact, in the first iteration itself..

Rearranging, #2^x=(6-2^(1+1/sqrt x))#.

Equating logarithms,

#x= ln(6-2^(1+1/sqrt x))/ln 2#

Choosing the discrete analogue

#x_n=ln(6-2^(1+1/sqrt x_(n-1)))/ln 2, n = 1, 2, 3, ..#,

with the guess solution 1 as the starter #x_0=1#,

the first iteration, in double precision mode, declares

#x_1=.1000000000000000e+00# and

#x_1-x_0=.0000000000000000e+00#

Now, substitution x = 1 shows that 1 is the solution, in exactitude..

Considering #f(x) = f_1(x)+f_2(x)# with

#f_1(x)=2^x# and #f_2(x) = 2^(1+1/sqrt(x))# we have that in the range #x > 0, x in RR#

#f_1(x)# is analytic nonlinear strictly increasing and #f_2(x)# is analytic nonlinear strictly decreasing so their sum must have an unique minimum at #x_0#. Both functions are analytic in this range so the minimum obeys the condition

#f'(x_0)=f'_1(x_0)+f'_2(x_0) = 0# or

#2^x - 2^(1/sqrt[x])/x^(3/2) =0#. But #x_0=1# obeys this condition so the only minimum is #f(x_0) = 6# located at #x_0=1#

Now, considering

#g(x) = f(x)-6# the only solution for #g(x)=0# is obviously #x=1#