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

((√2)−1)^x +((√2)+1)^x =((√6))^x

Commented byDwaipayan Shikari last updated on 04/Jul/20

x=2

Commented bydw last updated on 04/Jul/20

solution step by step please!

Answered by 1549442205 last updated on 05/Jul/20

⇔((((√2)−1)/(√6)))^x +((((√2)+1)/(√6)))^x =1 For x=2 we get ((((√2)−1)/(√6)))^2 +((((√2)+1)/(√6)))^2 = ((3−2(√2)+3+2(√2))/6)=1 ,so x=2 is the root of given equation.We prove that it is the unique root. Putting f(x)=((((√2)−1)/(√6)))^x +((((√2)+1)/(√6)))^x We have f ′(x)=⇔((((√2)−1)/(√6)))^x ln((((√2)−1)/(√6)))+((((√2)+1)/(√6)))^x ln((((√2)+1)/(√6)))<0 because (((√2)±1)/(√6))<1⇒ln((((√2)±1)/(√6)))<0 ⇒f(x) is an decreasing function on (−∞;+∞),so the finded root x=2 is unique

Commented bydw last updated on 04/Jul/20

Thank you!

Commented by1549442205 last updated on 05/Jul/20

You are welcome sir