Question

Se log2= 0,3 e log3= 0,48
determine:
a) Log12=
b) Log0,04=
c) Log³² na base 9=

Answer 1

a) Log 12 = log 2*3*2 = 2log2 + log 3 = 2.0,3 + 0,48 = 1,08

b) Log 0,04 = log 4/100 = log 4 - log 100 = log 2.2 - log 10.10 = 2log2 - 2log 10 = 2(log 2 - log 10) = 2(0,3-1) = -2.0,7 = -1,4

c) Log₉ 32 = log 32/log 9 = log 2⁵/ log 3² = 5/2(log 2 / log 3) = 2,5.(0,3/0,48) = 2,5.0,625 = 1,5625