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

Sep 17, 2016

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, ${2}^{x} = \left(6 - {2}^{1 + \frac{1}{\sqrt{x}}}\right)$.

Equating logarithms,

$x = \ln \frac{6 - {2}^{1 + \frac{1}{\sqrt{x}}}}{\ln} 2$

Choosing the discrete analogue

${x}_{n} = \ln \frac{6 - {2}^{1 + \frac{1}{\sqrt{x}} _ \left(n - 1\right)}}{\ln} 2 , n = 1 , 2 , 3 , . .$,

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..

Sep 17, 2016

The only solution is $x = 1$

#### Explanation:

Considering $f \left(x\right) = {f}_{1} \left(x\right) + {f}_{2} \left(x\right)$ with

${f}_{1} \left(x\right) = {2}^{x}$ and ${f}_{2} \left(x\right) = {2}^{1 + \frac{1}{\sqrt{x}}}$ we have that in the range $x > 0 , x \in \mathbb{R}$

${f}_{1} \left(x\right)$ is analytic nonlinear strictly increasing and ${f}_{2} \left(x\right)$ 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

$f ' \left({x}_{0}\right) = f {'}_{1} \left({x}_{0}\right) + f {'}_{2} \left({x}_{0}\right) = 0$ or

${2}^{x} - {2}^{\frac{1}{\sqrt{x}}} / {x}^{\frac{3}{2}} = 0$. But ${x}_{0} = 1$ obeys this condition so the only minimum is $f \left({x}_{0}\right) = 6$ located at ${x}_{0} = 1$

Now, considering

$g \left(x\right) = f \left(x\right) - 6$ the only solution for $g \left(x\right) = 0$ is obviously $x = 1$