Question

sabendo que log 2= A e log de 3=B, calclule log (8. raiz quadrada de 27)

Answer 1

→ log(2) = A

→ log(3) = B

______________________

$log(8\sqrt{27} ) = log(8) + log(\sqrt{27} ) \\\\ = log(2^3) + log(\sqrt{3^3}) \\\\ = 3log(2) + log(3^{3 \over 2}) \\\\ = 3A + {3 \over 2}log(3) \\\\ \boxed{ = 3A + {3B \over 2}}$