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(sqrt(5)+2)^(x-1)>=(sqrt(5)-2)^((x-1)/(x+1)) desigualdades

En la desigualdad la incógnita

Solución

Ha introducido [src]
                               x - 1
                               -----
           x - 1               x + 1
/  ___    \         /  ___    \     
\\/ 5  + 2/      >= \\/ 5  - 2/     
$$\left(2 + \sqrt{5}\right)^{x - 1} \geq \left(-2 + \sqrt{5}\right)^{\frac{x - 1}{x + 1}}$$
(2 + sqrt(5))^(x - 1) >= (-2 + sqrt(5))^((x - 1)/(x + 1))
Respuesta rápida [src]
  /                        /        /       ___\             \\
  |                        |     log\-2 + \/ 5 /             ||
Or|And(1 <= x, x < oo), And|-1 + --------------- <= x, x < -1||
  |                        |         /      ___\             ||
  \                        \      log\2 + \/ 5 /             //
$$\left(1 \leq x \wedge x < \infty\right) \vee \left(-1 + \frac{\log{\left(-2 + \sqrt{5} \right)}}{\log{\left(2 + \sqrt{5} \right)}} \leq x \wedge x < -1\right)$$
((1 <= x)∧(x < oo))∨((x < -1)∧(-1 + log(-2 + sqrt(5))/log(2 + sqrt(5)) <= x))
Respuesta rápida 2 [src]
         /       ___\               
      log\-2 + \/ 5 /               
[-1 + ---------------, -1) U [1, oo)
          /      ___\               
       log\2 + \/ 5 /               
$$x\ in\ \left[-1 + \frac{\log{\left(-2 + \sqrt{5} \right)}}{\log{\left(2 + \sqrt{5} \right)}}, -1\right) \cup \left[1, \infty\right)$$
x in Union(Interval(1, oo), Interval.Ropen(-1 + log(-2 + sqrt(5))/log(2 + sqrt(5)), -1))