Question

$log_5x-log_x25=1$

Answer 1

ОДЗ
x > 0 , x ≠ 1

log(5, x) - (log 5, 25) / (log 5, x) = 1 
log(5, x) -  2 /log(5, x) = 1
log(5, x)^2 - log(5,x) - 2 = 0 
t^2  - t - 2 = 0 
D = 1 + 4*2 = 9
t1 = (1 + 3)/2 = 2
t2  =( 1 - 3)/2 = - 1

log (5, x) = 2 
x = 5^2
x = 25

log (5,x) - 1
x = 5^(-1)
x = 1/5
x = 0,2

Ответ
25;  0,2