Question

Se log 2 = 0,30 e log 3 = 0,48, calcular:
a) log 45
b) log 2,4

Answer 1

Fatore o 45:

45 | 5
9 | 3
3 | 3
1

$log45 \\\\ log 5 \cdot 3^{2} \\\\ log\frac{10}{2}+log3^{2} \\\\ log10-log2+2 \cdot log3 \\\\ 1-0,3+2 \cdot 0,48 \\\\ 1-0,3+0,96 = \boxed{\boxed{1,66}}$


b)
$log2,4 \\\\ log\frac{24}{10} \\\\ log\frac{12}{5} \\\\ log12-log5 \\\\ log3 \cdot 2^{2}-log5 \\\\ log3+2 \cdot log2-log5 \\\\ 0,48+2 \cdot (0,3)-0,7 \\\\ 0,48+0,6-0,7 = \boxed{\boxed{0,38}}$

Answer 2

a) Log 45

= Log 3² * 5

= 2 Log 3 + Log 5

= 2 * 0,48 + 0,69

= 0,96 + 0,69

= 1,65

b) Log 2,4

= Log 24/10

= Log 24 - Log 10

= Log 2³ * 3 - 1

= 3 Log 2 + Log 3 - 1

= 3 * 0,30 + 0,48 - 1

= 0,9 + 0,48 - 1

= 1,38 - 1

= 0,38