Question

07-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) 120

Answer 1

Explicação passo-a-passo:

a) Lembre-se que:

$\text{log}_{a}~(b\cdot c)=\text{log}_{a}~b+\text{log}_{a}~c$

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

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

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

$\text{log}~6=0,778$

b) Lembre-se que:

$\text{log}_{a}~b^{n}=n\cdot\text{log}_{a}~b$

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

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

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

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

$\text{log}~18=1,255$

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

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

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

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

$\text{log}~50=1,699$

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

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

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

$\text{log}~32=1,204$

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

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

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

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

$\text{log}~120=2,079$