Question

Como calcular A=log1024 na base 2 + log625 na base 1/3 ?

Answer 1

 $A=log_{2} 1024 + log_{1/3}625=10 + \frac{log625}{log(1/3)} =10 + \frac{2,7959}{-0,4771} =$
$=10-5,8599=4,1401$