Impartim cu [tex]25^x[/tex] :
[tex]\left(\dfrac{4}{25}\right)^x+\left(\dfrac{10}{25}\right)^x=1\\ \\ \\\Rightarrow \left(\dfrac{4}{25}\right)^x+\left(\dfrac{2}{5}\right)^x=1\\ \\ \\ \Rightarrow \left(\dfrac{2}{5}\right)^{2x}+\left(\dfrac{2}{5}\right)^x=1[/tex]
Facem notatia: [tex]t=\left(\dfrac{2}{5}\right)^x, \ \ \ t>0.[/tex]
Rescriem: [tex]t^2+t=1[/tex]
Rearanjam si rezolvam ecuatia de grad 2:
[tex]t^2+t-1=0\\ \\ \Delta=1+4=5\\ \\ \\ t_1=\dfrac{-1+\sqrt{5}}{2}>0\\ \\ t_2=\dfrac{-1-\sqrt{5}}{2}<0.[/tex]
A doua solutie pica, pentru ca e negativa. Asadar, avem:
[tex]t=\dfrac{\sqrt{5}-1}{2}\\ \\ \left(\dfrac{2}{5}\right)^x=\dfrac{\sqrt{5}-1}{2}\\ \\ \\ x=\log_{\frac{2}{5}}\left(\dfrac{\sqrt{5}-1}{2}\right).[/tex]
Se mai poate aranja daca vrei sa arate mai frumos...
