Question

[sqrt(2)+1]^x + [sqrt(2)-1]^x = 6

Answer 1

$( \sqrt{2}+1)^x +( \sqrt{2} -1)^x=6 \\ fie :( \sqrt{2}+1)^x=t;t\ \textgreater \ 0 \\ t+ \frac{1}{t}=6 \\ t^2-6t+1=0$
Δ=36-4=32
t1=(6-4√2)/2=3-2√2
t2=(6+4√2)/2=3+2√2

revenim la substitutie
(√2+1)^x=(3+2√2) ⇔(√2+1)^x=(√2+1)² ⇔ x=2
(√2+1)^x=(3-2√2) ⇔(√2+1)^x=(√2+1)^(-2)⇔x=-2