Question

Sa se determine multimea tuturor valorilor lui x ∈ R stiind ca numerele
$9^{x}-1$ , $6^{x}$ , $4^{x} + 1$ sunt in progresie aritmetica in aceasta ordine.

Answer 1

Răspuns:

x=0.

Explicație pas cu pas:

Daca a,b,c sunt 3 termeni consecutivi a unei progresii aritmetice, atunci a+c=2·b

$Deci~9^{x}-1+4^{x}+1=2*6^{x},~9^{x}+4^{x}=2*6^{x} |:4^{x},~\frac{9^{x}}{4^{x}} +\frac{4^{x}}{4^{x}}=2*\frac{6^{x}}{4^{x}},~(\frac{9}{4})^{x}+1=2*(\frac{6}{4})^{x} ,~(\frac{9}{4})^{x}+1=2*(\frac{3}{2})^{x},~deoarece~ (\frac{9}{4})^{x}=((\frac{3}{2})^{2})^{x}= ((\frac{3}{2})^{x})^{2},~notam~(\frac{3}{2})^{x}=y,~si~obtinem\\y^{2}+1=2y,~y^{2}-2y+1=0,~(y-1)^{2}=0,~deci~y-1=0,~y=1.~Atunci~obtinem~(\frac{3}{2})^{x}=1,~deci~x=0.$