Kaj je x, če je log_4 x = 1/2 + log_4 (x-1)?

Odgovor:

# x = 2 #

Pojasnilo:

Kot # log_4 x = 1/2 + log_4 (x-1) #

# log_4x-log_4 (x-1) = 1/2 #

ali # log_4 (x / (x-1)) = 1/2 #

t.j. # x / (x-1) = 4 ^ (1/2) = 2 #

in # x = 2x-2 #

t.j. # x = 2 #

Odgovor:

# x = 2 #.

Pojasnilo:

# log_4x = 1/2 + log_4 (x-1) #.

#:. log_4 x-log_x (x-1) = 1/2 #.

#:. log_4 {x / (x-1)} = 1/2 … ker, log_bm-log_bn = log_b (m / n) #.

#:. {x / (x-1)} = 4 ^ (1/2) = 2, … ker, "definicija" log #.

#:. x = 2 (x-1) = 2x-2 #.

#:. -x = -2, ali, x = 2 #.

To root izpolnjuje eqn.

#:. x = 2 #.