Question

$4^{x} + 10^{x} = 25^{x}$ . Cum se rezolva ec. exponentiala ???

Answer 1

Impartim cu $25^x$ :

$\left(\dfrac{4}{25}\right)^x+\left(\dfrac{10}{25}\right)^x=1\\ \\ \\\Rightarrow \left(\dfrac{4}{25}\right)^x+\left(\dfrac{2}{5}\right)^x=1\\ \\ \\ \Rightarrow \left(\dfrac{2}{5}\right)^{2x}+\left(\dfrac{2}{5}\right)^x=1$

Facem notatia: $t=\left(\dfrac{2}{5}\right)^x, \ \ \ t>0.$

Rescriem: $t^2+t=1$

Rearanjam si rezolvam ecuatia de grad 2:

$t^2+t-1=0\\ \\ \Delta=1+4=5\\ \\ \\ t_1=\dfrac{-1+\sqrt{5}}{2}>0\\ \\ t_2=\dfrac{-1-\sqrt{5}}{2}<0.$

A doua solutie pica, pentru ca e negativa. Asadar, avem:

$t=\dfrac{\sqrt{5}-1}{2}\\ \\ \left(\dfrac{2}{5}\right)^x=\dfrac{\sqrt{5}-1}{2}\\ \\ \\ x=\log_{\frac{2}{5}}\left(\dfrac{\sqrt{5}-1}{2}\right).$

Se mai poate aranja daca vrei sa arate mai frumos...