Let a, b, and c be the roots of \(24x^3 - 121x^2 + 87x - 8 = 0.\)Find \(\log_3(a)+\log_3(b)+\log_3(c)\)
im stuck on this problem, and my parents forgot how to do logs😂...
any help would be appreciated! thank you so much!
sum of roots is 121/24
\(a+b+c=\frac{121}{24}\\ ab+ac+bc=\frac{87}{24}=\frac{29}{8}\\ abc=\frac{8}{24}=\frac{1}{3}\)
\(\log_3(a)+\log_3(b)+\log_3(c)\\ =log_3(abc)\\ =log_3(\frac{1}{3})\\ =log_3(3^{-1})\\ =-1log_3(3)\\ =-1\)
