Solve-for-x-1-2-x-1-2-x-14-

Question Number 217448 by ArshadS last updated on 14/Mar/25

S o l v e f o r x :

{(1 + \sqrt{2})}^{x} + {(1 - \sqrt{2})}^{x} = 14

Commented by Frix last updated on 14/Mar/25

a_{n} = {(1 + \sqrt{2})}^{n} + {(1 - \sqrt{2})}^{n}; n \in N

can be written as

a_{1} = 2

a_{2} = 2

a_{n} = a_{n - 2} + 2 a_{n - 1} \forall n ⩾ 3

\Rightarrow

a_{3} = 6

a_{4} = 14

a_{5} = 34

\dots

Commented by Hanuda354 last updated on 15/Mar/25

a_{1} = 2

a_{2} = 6

a_{3} = 14 = 2 + 2 \cdot 6

a_{4} = 34 = 6 + 2 \cdot 14

a_{5} = 82 = 14 + 2 \cdot 34

⋮

a_{n} = a_{n - 2} + 2 a_{n - 1} for \forall n ⩾ 3

{(1 + \sqrt{2})}^{3} + {(1 - \sqrt{2})}^{3} = 14

Answered by MrGaster last updated on 14/Mar/25

(1) : Let a = 1 + \sqrt{2} \land b = 1 - \sqrt{2}, b = - \frac{1}{a}

a^{x} + b^{x} = 14

for x = 3 :

a^{3} + b^{3} = {(1 + \sqrt{2})}^{3} + {(1 - \sqrt{2})}^{3}

{(1 + \sqrt{2})}^{3} = 7 + 5 \sqrt{2}

({(1 - \sqrt{2})}^{3} = 7 - 5 \sqrt{2}

(7 + 5 \sqrt{2}) + (7 + 5 \sqrt{2}) = 14

∴ x = 3

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

∵∣ 1 - \sqrt{2} ∣< 1, The term {(1 - \sqrt{2})}^{x} decaysi

exponentally ., {(1 + \sqrt{2})}^{x} Exponential growth, f (x) for x > 0 Is strictly incrementaln

ensurig a unique solution .

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

= {(\sqrt{2})}^{x} ({(\frac{2}{\sqrt{2}})}^{x} + 0^{x}) = (\sqrt{2}) ({(\sqrt{2})}^{x}) = 2^{x}

\Rightarrow 2^{x} = 14

\Rightarrow x = \log_{2} 14

x = \frac{\ln 14}{\ln 12}

x = \frac{\ln (2 \times 7)}{\ln 2} = \frac{\ln 2 + \ln 7}{\ln 2} = 1 + \frac{\ln 7}{\ln 2}

x = 1 + \log_{2} 7

x = 3 (since \log_{2} 7 \approx 2.807)

Answered by Rasheed.Sindhi last updated on 14/Mar/25

{(1 + \sqrt{2})}^{x} + {(1 - \sqrt{2})}^{x} = 14

A s s u m i n g x \in Z

\underset{y}{\underset{⏟}{{(\sqrt{2} + 1)}^{x}}} + {(- 1)}^{x} {(\sqrt{2} - 1)}^{x} = 14

y > 0

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

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

y + {(- 1)}^{x} (\frac{1}{y}) = 14

∙ I f x \in O

y - \frac{1}{y} = 14

y^{2} - 14 y - 1 = 0

y = \frac{14 \pm \sqrt{196 + 4}}{2} = 7 \pm 5 \sqrt{2}

∵ y > 0 ∴ y = 7 + 5 \sqrt{2}

{(\sqrt{2} + 1)}^{x} = 7 + 5 \sqrt{2}

x \log (\sqrt{2} + 1) = \log (7 + 5 \sqrt{2})

x = \frac{\log (7 + 5 \sqrt{2})}{\log (\sqrt{2} + 1)} = \log_{(\sqrt{2} + 1)} (7 + 5 \sqrt{2})

x = 3

∙ I f x \in E

y + \frac{1}{y} = 14

y^{2} - 14 y + 1 = 0

y = \frac{14 \pm \sqrt{196 - 4}}{2} = \frac{14 \pm 8 \sqrt{3}}{2} = 7 \pm 4 \sqrt{3}

{(\sqrt{2} + 1)}^{x} = 7 \pm 4 \sqrt{3}

x \log (\sqrt{2} + 1) = \log (7 \pm 4 \sqrt{3})

x = \frac{\log (7 \pm 4 \sqrt{3})}{\log (\sqrt{2} + 1)} = \log_{(\sqrt{2} + 1)} (7 \pm 4 \sqrt{3}) \notin Z

Answered by profcedricjunior last updated on 14/Mar/25

{(1 + \sqrt{2})}^{x} + {(1 - \sqrt{2})}^{x} = 14

=> {(1 + \sqrt{2})}^{2 x} + 1 - 14 {(1 + \sqrt{2})}^{x} = 0 \overset{1 + \sqrt{2} = t}{}

=> t^{2} - 14 t + 1 = 0 => {(t - 7)}^{2} - 48 = 0

=> (t - 7 - 4 \sqrt{3}) (t - 7 + 4 \sqrt{3}) = 0

=> t = 7 - 4 \sqrt{3} o u t = 7 + 4 \sqrt{3}

=> {\begin{cases} {(1 + \sqrt{2})}^{x} = 7 - 4 \sqrt{3} \\ {(1 + \sqrt{2})}^{x} = 7 + 4 \sqrt{3} \end{cases} => {\begin{cases} {(1 + \sqrt{2})}^{x} = {(2 - \sqrt{3})}^{2} \\ {(1 + \sqrt{2})}^{x} = {(2 + \sqrt{3})}^{2} \end{cases}

=> {\begin{cases} x = \frac{2 l n (2 - \sqrt{3})}{l n (1 + \sqrt{2})} = 2 {l o g}_{(1 + \sqrt{2})} (2 + \sqrt{3}) \\ x = \frac{2 l n (2 + \sqrt{3})}{l n (1 + \sqrt{2})} = 2 {l o g}_{(1 + \sqrt{2})} (2 + \sqrt{3}) \end{cases}

S_{R} = {2 {l o g}_{(1 + \sqrt{2})} (2 - \sqrt{3}), 2 {l o g}_{(1 + \sqrt{2})} (2 + \sqrt{3})}