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).
+26367 Let a,b, and c be the roots of \(24x^3 - 121x^2 + 87x - 8\) = 0
Find\( \log_3(a)+\log_3(b)+\log_3(c)\).
Vieta: \(-abc = -\dfrac{8}{24}\)
\(\begin{array}{|rcll|} \hline && \mathbf{\log_3(a)+\log_3(b)+\log_3(c)} \\\\ &=& \log_3(abc) \\\\ &=& \dfrac{\log(abc)}{\log(3)} \quad | \quad -\dfrac{8}{24}=-abc \\\\ &=& \dfrac{\log \left(\dfrac{8}{24} \right)}{\log(3)} \\\\ &=& \dfrac{\log \left(\dfrac{1}{3} \right)}{\log(3)} \\\\ &=& \dfrac{ \log(1)-\log(3) }{\log(3)} \quad | \quad \log(1) = 0\\\\ &=& \dfrac{ -\log(3) }{\log(3)} \\\\ &=& \mathbf{-1} \\ \hline \end{array}\)
