$\displaystyle log_b a=x$
$\displaystyle b^x=a$
em que
b = base
a =logaritmando
x= logaritmo
Atenção para a condição de existência dos logaritmos, precisamos que a>0 e b>0 e b≠1.
Caso deseje revisar as propriedade clique aqui
1) Calcule os seguintes logaritmos:
$\displaystyle a)log_2 128=$
$\displaystyle log_2 128=log_2 2^7=x$
$\displaystyle 2^x=2^7\Leftrightarrow x=7$
$\displaystyle b)log_2 16=$
$\displaystyle log_2 16=log_2 2^4=x$
$\displaystyle 2^x=2^4\Leftrightarrow x=4$
$\displaystyle c) log3 27 =$
$\displaystyle log_3 27=log_3 3^3=x$
$\displaystyle 3^x=3^3\Leftrightarrow x=3$
$\displaystyle d) log_{10} 10=$
$\displaystyle 10^x=10^1\Leftrightarrow x=1$
$\displaystyle e) log 1000=$
$\displaystyle log 1000=log 10^3=x$
$\displaystyle 10^x=10^3\Leftrightarrow x=3$
$\displaystyle f) log_2 \frac{1}{4}=$
$\displaystyle log_2 \frac{1}{4}=log_2 \frac{1}{2^2}=log_2 2^{-2}=x$
$\displaystyle 2^x=2^{-2}\Leftrightarrow x=-2$
$\displaystyle g) log_5 625=$
$\displaystyle log_5 625=log_5 5^4=x$
$\displaystyle 5^x=5^4\Leftrightarrow x=4$
$\displaystyle h)log_\frac{1}{2} 64=$
$\displaystyle log_\frac{1}{2} 64=log_\frac{1}{2} 2^6=x$
$\displaystyle \left (\frac{1}{2} \right )^x=2^6\Leftrightarrow 2^{-1}x=6\Leftrightarrow -x=6 \Leftrightarrow x=-6$
$\displaystyle i) log_1 7=$
Não satisfaz a condição de existência, b tem que ser diferente de zero.
$\displaystyle j) log_{20} 40=$
$\displaystyle log_{20}40={log_{20}2\cdot 20}={log_{20}2+log_{20}20}={log_{20}2}+1=1+log_{20}2$
$\displaystyle log_{20}2\approx 0,22$
$\displaystyle log_{20} 40=1 + 0,22=1,22$
2) Se log 2 = 0,30 e log 3 = 0,48, calcule:
a) log 6
$\displaystyle log 2\cdot3=log2+log3=0,30+0,48 =0,78$
b) log 12
$\displaystyle log 2^2\cdot3=2log2+log3=2\cdot0,30+0,48 =0,60+0,48=1,08$
c) log 9
$\displaystyle log 3^2=2log3=2\cdot0,48=0,96$
d) log 24
$\displaystyle log 2^3\cdot3=3log2+log3=3\cdot0,30+0,48 =0,90+0,48=1,38$
e) log 18
$\displaystyle log 2\cdot3^2=log2+2log3=0,30+2\cdot0,48 =0,30+0,96=1,26$
f) log 200
$\displaystyle log 2\cdot10^2=log2+2log10=0,30+2\cdot1 =2,30$
g) log 50
$\displaystyle log 5\cdot10=log5+1$(I)
$\displaystyle log 5=log\frac{10}{2}=log10 - log 2 = 1 - 0,30 = 0,70$
Retomando (I) temos
$\displaystyle log5+1=0,70 + 1= 1,70 $
h) log 1,5
$\displaystyle log 1,5=log\frac{3}{2}=log3-log2=0,48-0,30=0,18$
i) log 16
$\displaystyle log 2^4=4log2=4\cdot0,30=1,20$
j)$\displaystyle log_3 2$
$\displaystyle log_3 2=\frac{log3}{log2}=\frac{0,48}{0,30}=\frac{4,8}{3}=1,6$
3) Usando apenas que log 20 = 1, 30, log 30 = 1, 47 e log 60 = 1, 79, calcule:
a) log 4 + log 5
$\displaystyle log4+log5=log4\cdot 5=log20=1,30$
b) log 5 + log 6
$\displaystyle log 5+log6=log5\cdot 6=log30 = 1,47$
c) log 150 − log 5
$\displaystyle log 150-log5=log\frac{150}{5}=log30=1,47$
d) log 120 − log 6
$\displaystyle log 120-log6=log\frac{120}{6}=log20=1,30$
e) log 5 + log 12
$\displaystyle log 5+log12=log5\cdot12=log60=1,79$
f) log 180 − log 9
$\displaystyle log 180-log9=log\frac{180}{9}=log20=1,30$
g) log 16 + 2 log 5
$\displaystyle log 16+2log5=log16+log5^2=log16+log25=log16\cdot25=log400=log20^2=$
$\displaystyle =2log20=2\cdot1,30=2,60$
h) log 600 − log 30
$\displaystyle log 600-log30=log\frac{600}{30}=log20=1,30$
i) 2 log 15 + log 4
$\displaystyle 2 log 15+log4 = log 15^2+log4=log225+log4=log225\cdot4=log900=$
$\displaystyle =log30^2=2log30=2\cdot 1,47=2,94$
j) log 1800 − log 60
$\displaystyle log 1800-log60=log\frac{1800}{60}=log30=1,47$
k) log 225 + 2 log 4
$\displaystyle log 225+2 log4=log225+log4^2=log225+log16=$
$\displaystyle log225\cdot16=log3600=log60^2=2log60=2\cdot1,79=3,58$
l) log 1200 − log 20
$\displaystyle log 1200-log20=log\frac{1200}{20}=log60=1,79$
m) log 15 + log 40
$\displaystyle log 15+log40=log15\cdot40=log600=log30 +log20 = 1,47+1,30=2,77$
n) log 1800 − log 30
$\displaystyle log1800-log30=log\frac{1800}{30}=log60=1,79$
o) log 20 + log 45
$\displaystyle log20+log45=log20\cdot45=log900=log30^2=2log30=2\cdot1,47=2,94$
p) log 6000 − log 30
$\displaystyle log6000-log30=log\frac{6000}{30}=log200=log20\cdot10=log20+log10=1,30+1=2,30 $
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