Question

se log2=0,3 e log3 =0,48, calcule log 12, log13,5 ,log 18

Answer 1

$log12=log (2^{2}*3)=log 2^{2}+log3=2log2+log3$
$=2*0,3+0,48=1,08$

$log13,5=log \frac{135}{10} =log \frac{ 3^{3} *5}{10} =log 3^{3} +log5-log10$
$=3log3+log \frac{10}{2} -log10=3log3+(log10-log2)-log10$
$=3*0,48+(1-0,3)-1=1,44+0,7-1=1,14$

$log18=log( 3^{2}*2 )=2log3+log2=2*0,48+0,3$
$=0,96+0,3=1,26$