Question

Numerele reale strict pozitive x si y verifică relațiile:
$log_{4}(x) = log_{6}(y) = log_{9}(x + y)$
Cât este
$\frac{y}{x}$
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Answer 1

$log_4x-log_6y=log_9(x+y)=k\\ \\ \Rightarrow x=4^k \\ \\ \Rightarrow y=6^k\\ \\ \Rightarrow x+y=9^k\\ \\ \Rightarrow 9^k\cdot 4^k=(6^k)^2 \\ \\ \Rightarrow (x+y)\cdot x=y^2 \\ \\ OBS: \ Din \ conditiile \ de \ existenta \ a \ logaritmilor \ avem: \ x,y>0\\ \\ x^2+xy=y^2 \ |:y^2 \Rightarrow (\frac{x}{y})^2+\frac{x}{y}=1\\ \\ (\frac{x}{y})^2+\frac{x}{y}-1=0\\ \\ \Delta=1^2-4\cdot 1\cdot (-1)=5\\ \\ \frac{x}{y}_{1,2}=\frac{-1 \pm \sqrt{5}}{2}\\ \\ \frac{x}{y}_1=\frac{-1-\sqrt{5}}{2}<0, \ nu \ convine \ deoarece \ x \ si \ y \ sunt \ numere \ nenegative.\\ \\ \frac{x}{y}_2=\frac{-1+\sqrt{5}}{2}>0 \\ \\ \frac{y}{x}=\frac{2}{-1+\sqrt{5}}=\frac{2(-1-\sqrt{5})}{(-1)^2-(\sqrt{5})^2}=\frac{-2(1+\sqrt{5})}{-4}=\frac{1+\sqrt{5}}{2}\\ \\ RASPUNS: \ \frac{y}{x}=\frac{1+\sqrt{5}}{2}$