Question

Dados log 2 = 0,3010,log 3 = 0,4771 e log 5 = 0,6990,calcule todos abaixo:
A)log 72
B)log raiz quadrada de 5
C)log 25
D)log 50

Answer 1

a) $\log_{10}[72] = \log_{10}[2 \cdot 36] = \log_{10}[2 \cdot 6^2] = \log_{10}[2] + \log_{10}[6^2] = \log_{10}[2] + 2 \cdot \log_{10}[6] = \log_{10}[2] + 2 \cdot \log_{10}[3 \cdot 2] = \log_{10}[2] + 2 \cdot ( \log_{10}[3] + \log_{10}[2])$

$\log_{10}[72] = 0,3010 + 2 \cdot (0,4771 + 0,3010)$

$\log_{10}[72] = 0,3010 + 2 \cdot 0,7781$

$\log_{10}[72] = 0,3010 + 1,5562$

$\log_{10}[72] = 1,8562$

b)$\log_{10}[\sqrt{5}] = log_{10}[5^{\frac{1}{2}}] = \dfrac{1}{2} \cdot \log_{10}[5]$

$\log_{10}[\sqrt{5}] = \dfrac{1}{2} \cdot 0,6990$

$\log_{10}[\sqrt{5}] = 0,3495$

c) $\log_{10}[25] =  \log_{10}[5^2] = 2 \cdot \log_{10}[5}$

$\log_{10}[25] = 2 \cdot 0,6990$

$\log_{10}[25] = 1,3980$

d) $\log_{10}[50] = \log_{10}[5 \cdot 10] = \log_{10}[5 \cdot 2 \cdot 5] = 2 \cdot \log_{10}[5] + \log_{10}[2]$

d) $\log_{10}[50] = 2 \cdot 0,6990 + 0,3010$

$\log_{10}[50] = 1,3980 + 0,3010$

$\log_{10}[50] = 1,3980 + 0,3010$

$\log_{10}[50] = 1,6990$