Question

log_3 (27√3) + log_2 (∛16) =

Answer 1

$log_3(27\sqrt3)+log_2\sqrt[3]{16}=log_3\, 3^{7/2}+log_22^{4/3}=\frac{7}{2}+\frac{4}{3}=\frac{29}{6}$

Answer 2

log₃ (27√3) + log₂(∛16) =log₃(3³3¹/²))+log₂(2⁴/³)=log₃(3⁷/²)+(4/3)=7/2+4/3=

=(21+8)/6=29/6=4  5/6