Sabendo que log2=0,3 e log3=0,48, calcule:
log de 50 na base 9
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Log(9)50 =
Log(3²)(5*10) =
1/2*Log(3)5+1/2*Log(3)10 =
1/2*Log(3)(10/2)+1/2*1/(Log(10)3) =
1/2*Log(3)10 - Log(3)2 + 1/2*(1/0,48) =
1/2*1/(log(10)3) - (Log 3)/(Log 2) + 1/2*(100/48) =
1/2*1/0,48 - 0,48/0,3+1/2*(25/12) =
1/2*100/48 - 48/3+25/24 =
100/96-16+25/24 =
50/48 - 16/1 + 25/24 =
mmc (48,1,24) = 48
(48:48*50)/48 - (48:1*16)/48+(48:24*25)/48 =
50/48 - (48*16)/48 + (2*25)/48 =
50/48 - 3/48 + 50/48 =
(50-3+50)/48 =
103/48
Log(3²)(5*10) =
1/2*Log(3)5+1/2*Log(3)10 =
1/2*Log(3)(10/2)+1/2*1/(Log(10)3) =
1/2*Log(3)10 - Log(3)2 + 1/2*(1/0,48) =
1/2*1/(log(10)3) - (Log 3)/(Log 2) + 1/2*(100/48) =
1/2*1/0,48 - 0,48/0,3+1/2*(25/12) =
1/2*100/48 - 48/3+25/24 =
100/96-16+25/24 =
50/48 - 16/1 + 25/24 =
mmc (48,1,24) = 48
(48:48*50)/48 - (48:1*16)/48+(48:24*25)/48 =
50/48 - (48*16)/48 + (2*25)/48 =
50/48 - 3/48 + 50/48 =
(50-3+50)/48 =
103/48
lucaseich1:
Obrigado!
Observei mais uma vez a resposta,na penúltima linha... (50-3+50) = (50+50-3) = 100-3 = 97, logo a resposta será 97/48
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