Question

1)considerando as propriedades log2 = 0,3 e log3=0,48, calcule o valor de:
a)log(2⁴.3⁵)
b)log(⁵√90)
c)log 0,05
d)log 3,6
e)log 2000
2)admitindo que log 8 = p, obtenha, em função de p:
a)log 16
b)log 1,28

Answer 1

a)

$log(2^{4}*3^{5})=log(2^{4})+log(3^{5})$
$log(2^{4}*3^{5})=4*log(2)+5*log(3)$
$log(2^{4}*3^{5})=4*0,3 + 5*0,48$
$log(2^{4}*3^{5})=1,2+2,4$
$log(2^{4}*3^{5})=3,6$

b)

$log(\sqrt[5]{90})=log(\sqrt[5]{3^{2}*10})$
$log(\sqrt[5]{90})=log([{3^{2}*10}]^{1/5})$
$log(\sqrt[5]{90})=log(3^{2/5}*10^{1/5})$
$log(\sqrt[5]{90})=log(3^{2/5})+log(10^{1/5})$
$log(\sqrt[5]{90})=(2/5)*log(3)+(1/5)*log(10)$
$log(\sqrt[5]{90})=(2/5)*0,48+(1/5)*1$
$log(\sqrt[5]{90})=(0,96/5)+(1/5)$
$log(\sqrt[5]{90})=(0,96+1)/5$
$log(\sqrt[5]{90})=1,96/5$
$log(\sqrt[5]{90})=0,392$

c)

$log(0,05)=log(5/100)$
$log(0,05)=log(1/20)$
$log(0,05)=log(1)-log(20)$
$log(0,05)=0-log(2*10)$
$log(0,05)=-[log(2)+log(10)]$
$log(0,05)=-[0,3+1]$
$log(0,05)=-[1,3]$
$log(0,05)=-1,3$

d)

$log(3,6)=log(36/10)$
$log(3,6)=log(3^{2}*2^{2}/10)$
$log(3,6)=log(3^{2})+log(2^{2})-log(10)$
$log(3,6)=2*log(3)+2*log(2)-1$
$log(3,6)=2*0,48+2*0,3-1$
$log(3,6)=0,96+0,6-1$
$log(3,6)=0,56$

e)

$log(2000)=log(2*10^{3})$
$log(2000)=log(2)+log(10^{3})$
$log(2000)=0,3+3*log(10)$
$log(2000)=0,3+3*1$
$log(2000)=0,3+3$
$log(2000)=3,3$
__________________________________

$log(8)=p$
$log(2^{3})=p$
$3*log(2)=p$
$log(2)=p/3$

a)

$log(16)=log(2^{4})$
$log(16)=4*log(2)$
$log(16)=4*p/3$
$log(16)=4p/3$

b)

$log(1,28)=log(2^{7}/100)$
$log(1,28)=log(2^{7})-log(100)$
$log(1,28)=7*log(2)-log(10^{2})$
$log(1,28)=7*(p/3) - 2*log(10)$
$log(1,28)=(7p/3)-2*1$
$log(1,28)=(7p/3)-2$
$log(1,28)=(7p - 6) / 3$