Question

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

Answer 1

1

Explanation:

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..

Answer 2

The only solution is `x = 1`

Explanation:

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`