2-1-x-2-1-x-6-x-

Question Number 101803 by dw last updated on 04/Jul/20

{(\sqrt{2} - 1)}^{x} + {(\sqrt{2} + 1)}^{x} = {(\sqrt{6})}^{x}

Commented by Dwaipayan Shikari last updated on 04/Jul/20

x = 2

Commented by dw last updated on 04/Jul/20

s o l u t i o n s t e p b y s t e p p l e a s e!

Answered by 1549442205 last updated on 05/Jul/20

\Leftrightarrow {(\frac{\sqrt{2} - 1}{\sqrt{6}})}^{x} + {(\frac{\sqrt{2} + 1}{\sqrt{6}})}^{x} = 1

For x = 2 we get {(\frac{\sqrt{2} - 1}{\sqrt{6}})}^{2} + {(\frac{\sqrt{2} + 1}{\sqrt{6}})}^{2} =

\frac{3 - 2 \sqrt{2} + 3 + 2 \sqrt{2}}{6} = 1, so x = 2 is the root

of given equation . We prove that it is

the unique root .

Putting f (x) = {(\frac{\sqrt{2} - 1}{\sqrt{6}})}^{x} + {(\frac{\sqrt{2} + 1}{\sqrt{6}})}^{x}

We have f^{'} (x) =\Leftrightarrow {(\frac{\sqrt{2} - 1}{\sqrt{6}})}^{x} \ln (\frac{\sqrt{2} - 1}{\sqrt{6}}) + {(\frac{\sqrt{2} + 1}{\sqrt{6}})}^{x} \ln (\frac{\sqrt{2} + 1}{\sqrt{6}}) < 0

because \frac{\sqrt{2} \pm 1}{\sqrt{6}} < 1 \Rightarrow \ln (\frac{\sqrt{2} \pm 1}{\sqrt{6}}) < 0

\Rightarrow f (x) is an decreasing function on

(- \infty; + \infty), so the finded root x = 2 is unique

Commented by dw last updated on 04/Jul/20

T h a n k y o u!

Commented by 1549442205 last updated on 05/Jul/20

You are welcome sir