Odpowiedź:
Szczegółowe wyjaśnienie:
2 log 3 (x + 1) - log 3y - log₂8 = log (3 (x+1))² - log 3y - 3 = log 9 (x+1)² - log 3y - 3 = 2 - 1 - 3 = 1 - 3 = -2
[tex]10^{x}[/tex] = (3 (x+1))²
[tex]^{x}[/tex] = 2
[tex]10^{x}[/tex] = 3y
[tex]10^{x}[/tex] = (3y)¹
[tex]^{x}[/tex] = 1
