Question

Utilize as aproximações log 2 = 0,30 e log 3 = 0,48 para calcular:

a) log 72
b) log 14,4
c) log 1,296

Answer 1

Resposta:

a-) log 72 = log 3² . 2³

    log 72 = 2 log 3 + 3 log 2

    log 72 = 2 . 0,48 + 3.0,30

    log 72 = 0,96 + 0,90

    log 72 = 1,86

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b-) log 14,4 = log 144 / 10

    log 14,4 = log 2.72/10

    log 14,4 = log 2 + log 72 - log 10

    log 14,4 = 0,30 + 1,86 - 1

    log 14,4 = 1,16

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c-) log 1,296

    log 1,296 = log 1296/1000

    log 1,296 = log 72 . 18 / 1000

    log 1,296 = log 72 + log 18 - log 1000

    log 1,296 = log 72 + log 2 . 3² - log 1000

    log 1,296 = log 72 + log 2 + 2 log 3  - log 1000

    log 1,296 = 1,86 + 0,30 + 2.0,48 - 3

    log 1,296 = 1,86 + 0,30 + 0,96 - 3

    log 1,296 = 0,12

Answer 2

Resposta:

a)Resposta

$log72 = log9 + log8 = log {3}^{2} + log {2}^{3}$

$log {3}^{2} + log{2}^{3} = 2log3 + 3log2 = 1.86$

b)Resposta

$log14.4 = log \frac{144}{10}$

$log(144) - log(10) = log({12}^{2} ) - 1$

$2 log(12) - 1 = 2 log(6) + 2 log(2)$

$2 log(3) + 2log(2) + 2log(2) - 1 = 1.16$

c)Resposta

$log(1.296) = log( \frac{1296}{1000} ) = log(1296) - log(1000)$

$log(1296) - 3 = log( {6}^{4} ) - 3$

$= 4 log(3.2) - 3 = 4 log(3) + 4 log(2) - 3 = 0.12$