Question

Considere os seguintes dados log2=0,301, log3=0,477 e log5=0,699. Calcule:
a) log5 (base 12)
b) log4 (base 15)
c) log9 (base 60)
d) log15 (base 32)

Answer 1

Olá,

a) → $log_{12}5$

$\frac{log_{10}5 }{log_{10}12 }$

$\frac{log_{10}5 }{log_{10}(2X2X3) }$

$\frac{log_{10}5 }{log_{10}2+log_{10}2 +log_{10}3}$

$\frac{0,699}{0,301+0,301+0,477}$

$\frac{0,699}{1,079}$

0,647822057...

b) → $log_{15}4$

$\frac{log_{10}4 }{log_{10}15}$

$\frac{log_{10}(2X2) }{log_{10}(3X5)}$

$\frac{log_{10}2 +log_{10}2 }{log_{10} 3+log_{10}5 }$

$\frac{0,301+0,301}{0,477+0,699}$

$\frac{0,602}{1,176}$

0.5119047619...

c) → $log_{60} 9$

$\frac{log_{10}9 }{log_{10}60}$

$\frac{log_{10}(3X3) }{log_{10}(2X3X10)}$

$\frac{log_{10}3 +log_{10}3 }{log_{10} 2+log_{10}3+log_{10}10}$

$\frac{0,477+0,477}{0,301+0,477+1}$

$\frac{0,954}{1,778}$

0.53655793025...

d) → $log_{32}15$

$\frac{log_{10}15 }{log_{10}32}$

$\frac{log_{10}(3X5) }{log_{10}(2X2X2X2X2)}$

$\frac{log_{10}3 +log_{10}5 }{log_{10} 2+log_{10}2+log_{10}2+log_{10}2+log_{10}2}$

$\frac{0,477+0,699}{0,301+0,301+0,301+0,301+0,301}$

$\frac{1.176}{1,505}$

0.78139534883...

Espero ter te ajudado :D Bons estudos ^^