(log 13) (log x) (logₓy) = 2 Rešitev za y. ?

(log 13) (log x) (logₓy) = 2 Rešitev za y. ?
Anonim

Od # log_3 (13) = 1 / (log_13 (3)) #

imamo

# (log_3 (13)) (log_13 (x)) (log_x (y)) = (log_13 (x) / (log_13 (3))) (log_x (y)) #

Kvocient s skupno bazo 13 sledi spremembi osnovne formule, tako da

# log_13 (x) / (log_13 (3)) = log_3 (x) #, in

leva stran je enaka

# (log_3 (x)) (log_x (y)) #

Od

# log_3 (x) = 1 / (log_x (3)) #

leva stran je enaka

#log_x (y) / log_x (3) #

ki je sprememba osnove za

# log_3 (y) #

Zdaj, ko to vemo # log_3 (y) = 2 #, pretvorimo v eksponentno obliko, tako da

#y = 3 ^ 2 = 9 #.

Odgovor:

# y = 9 #

Pojasnilo:

Po uporabi #log_a (b) * dnevnik (b) _c = log_a (c) # identiteta, # log_3 (13) * log_13 (x) * log_x (y) = 2 #

# log_3 (x) * log_x (y) = 2 #

# log_3 (y) = 2 #

# y = 3 ^ 2 = 9 #