Объяснение:
log16(5x-14) = 0
5x-14 = 16^0 = 1
5x = 15
x = 3
log7(5-x) = log7(29-9x)
5-x = 29-9x
8x = 24
2log²25(x) + log25(x) - 1 = 0
Пусть, t = log25(x)
2t² + t - 1 = 0
D = 1+8 = 9
t1 = (-1+3)/4 = 0.5 ⇒ x = 25^(0.5) = √25 = 5
t2 = (-1-3)/4 = -1 ⇒ x = 25^(-1) = 1/25
Объяснение:
log16(5x-14) = 0
5x-14 = 16^0 = 1
5x = 15
x = 3
log7(5-x) = log7(29-9x)
5-x = 29-9x
8x = 24
x = 3
2log²25(x) + log25(x) - 1 = 0
Пусть, t = log25(x)
2t² + t - 1 = 0
D = 1+8 = 9
t1 = (-1+3)/4 = 0.5 ⇒ x = 25^(0.5) = √25 = 5
t2 = (-1-3)/4 = -1 ⇒ x = 25^(-1) = 1/25