Kaj je obratno y = -log (1.05x + 10 ^ -2)?

Odgovor:

# f ^ -1 (x) = (10 ^ -x-10 ^ -2) /1.05#

Pojasnilo:

Glede na: #f (x) = -log (1.05x + 10 ^ -2) #

Let #x = f ^ -1 (x) #

#f (f ^ -1 (x)) = -log (1.05f ^ -1 (x) + 10 ^ -2) #

Po definiciji #f (f ^ -1 (x)) = x #

#x = -log (1.05f ^ -1 (x) + 10 ^ -2) #

Pomnožite obe strani s -1:

# -x = log (1.05f ^ -1 (x) + 10 ^ -2) #

Na obeh straneh naj bo eksponent 10:

# 10 ^ -x = 10 ^ (log (1.05f ^ -1 (x) + 10 ^ -2)) #

Ker sta 10 in log inverzna, se desna stran zmanjša na argument:

# 10 ^ -x = 1.05f ^ -1 (x) + 10 ^ -2 #

Preklopite enačbo:

# 1.05f ^ -1 (x) + 10 ^ -2 = 10 ^ -x #

Odštejte 10 ^ -2 na obeh straneh:

# 1.05f ^ -1 (x) = 10 ^ -x-10 ^ -2 #

Obe strani delite s 1,05:

# f ^ -1 (x) = (10 ^ -x-10 ^ -2) /1.05#

Preverite:

#f (f ^ -1 (x)) = -log (1,05 ((10 ^ -x-10 ^ -2) / 1,05) + 10 ^ -2) #

#f (f ^ -1 (x)) = -log (10 ^ -x-10 ^ -2 + 10 ^ -2) #

#f (f ^ -1 (x)) = -log (10 ^ -x) #

#f (f ^ -1 (x)) = - (- x) #

#f (f ^ -1 (x)) = x #

# f ^ -1 (f (x)) = (10 ^ - (- log (1,05x + 10 ^ -2)) -10 ^ -2) /1.05#

# f ^ -1 (f (x)) = (10 ^ (log (1.05x + 10 ^ -2)) -10 ^ -2) /1.05#

# f ^ -1 (f (x)) = (1,05x + 10 ^ -2-10 ^ -2) /1.05

# f ^ -1 (f (x)) = (1,05x) /1,05

# f ^ -1 (f (x)) = x #

Oba pogoja preverita.