Question

Considerando log 2 = 0,301 ,
log 3 = 0,477 e log 5 = 0,699,calcule:
a) log 6

b) log 18

c) log 50

d) log 32

e) log 120

Answer 1

Resposta:

log 120=2,079

Explicação passo-a-passo:

a) Lembre-se que:

\text{log}_{a}~(b\cdot c)=\text{log}_{a}~b+\text{log}_{a}~cloga (b⋅c)=loga b+loga c

a) \text{log}~6=\text{log}~(2\cdot3)log 6=log (2⋅3)

\text{log}~6=\text{log}~2+\text{log}~3log 6=log 2+log 3

\text{log}~6=0,301+0,477log 6=0,301+0,477

\text{log}~6=0,778log 6=0,778

b) Lembre-se que:

\text{log}_{a}~b^{n}=n\cdot\text{log}_{a}~bloga bn=n⋅loga b

\text{log}~18=\text{log}~(2\cdot3^2)log 18=log (2⋅32)

\text{log}~18=\text{log}~2+2\cdot\text{log}~3log 18=log 2+2⋅log 3

\text{log}~18=0,301+2\cdot0,477log 18=0,301+2⋅0,477

\text{log}~18=0,301+0,954log 18=0,301+0,954

\text{log}~18=1,255log 18=1,255

c) \text{log}~50=\text{log}~(2\cdot5^2)log 50=log (2⋅52)

\text{log}~50=\text{log}~2+2\cdot\text{log}~5log 50=log 2+2⋅log 5

\text{log}~50=0,301+2\cdot0,699log 50=0,301+2⋅0,699

\text{log}~50=0,301+1,398log 50=0,301+1,398

\text{log}~50=1,699log 50=1,699

d) \text{log}~32=\text{log}~2^4log 32=log 24

\text{log}~32=4\cdot\text{log}~2log 32=4⋅log 2

\text{log}~32=4\cdot0,301log 32=4⋅0,301

\text{log}~32=1,204log 32=1,204

e) \text{log}~120=\text{log}~(2^3\cdot3\cdot5)log 120=log (23⋅3⋅5)

\text{log}~120=3\cdot\text{log}~2+\text{log}~3+\text{log}~5log 120=3⋅log 2+log 3+log 5

\text{log}~120=3\cdot0,301+0,477+0,699log 120=3⋅0,301+0,477+0,699

\text{log}~120=0,903+1,176log 120=0,903+1,176

\text{log}~120=2,079log 120=2,079