mira utiliza el cambio de bases para logaritmos para pasarlos a la misma base :)
por ejemplo pasémolas a base 3:
recuerda para ello lo que se hace es:
log (base 2/3) de (x-1) = log (base 3) (x-1) / log (base 3) de 2/3
luego queda la ecuación:
log (base 3) de (x-1)
----------------------------- - log (base 3) de (x-1) = 3
log (base 3) de 2/3
ahora podemos solucionar la ecuación
log₃ (x-1) - (log₃ 2/3)*(log₃ (x-1)) = 3*(log₃ 2/3)
los voy a combinar usando la propiedad de resta de logaritmos ya verás por qué:
(x-1)
log₃ ----------------------- = 3*(log₃ 2 - log₃3)
(x-1) ^ (log₃ 2/3)
(x-1)
log₃ ----------------------- = 3*(log₃ 2 - 1)
(x-1) ^ (log₃ 2 - 1)
(x-1)
log₃ ------------------------- = 3*(log₃ 2 - 1)
(x-1) ^ (log₃ 2)
-----------------
(x-1)
(x-1)(x-1)
log₃ ------------------------- = 3*(log₃ 2 - 1)
(x-1) ^ (log₃ 2)
(x-1)^2
log₃ ------------------------- = 3*(log₃ 2 - 1)
(x-1) ^ (log₃ 2)
(2-log₃2)log₃(x-1) = 3*(log₃ 2 - 1)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
(2-log₃2)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
(1 + 1- log₃2)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
(1 + log₃3 - log₃2)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(-1 - log₃3 + log₃2)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(-1 + log₃2 - log₃3)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(-1 + log₃2/3)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(log₃2/3 - 1)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(log₃2/3 - log₃3)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(log₃(2/3)/3)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(log₃(1/2))
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(log₃1 - log₃2)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
-(0 - log₃2)
3*(log₃ 2 - 1)
log₃(x-1) = ------------------------
log₃2
quizás desde aquí te des maña de hacerlo, me avisas si no :)
