👤
a fost răspuns

Calculați log in baza 2 din 3 ori log in baza 3 din 4 ori log in baza 4 din 5 ori...ori log in baza 7 din 8

Răspuns :

C04f
[tex]= \frac{lg3}{lg2}* \frac{lg4}{lg3}* \frac{lg5}{lg4}* \frac{lg6}{lg5}* \frac{lg7}{lg6}* \frac{lg8}{lg7}= \frac{lg8}{lg2}= log_{2}8= log_{2}2^3=3 [/tex]
[tex]\displaystyle log_23 \cdot log_34 \cdot log_45 \cdot ... \cdot log_78=log_23 \cdot log_34 \cdot log_45 \cdot log_56 \cdot log_67 \cdot log_78= \\ \\ = \frac{lg3}{lg2} \cdot \frac{lg4}{lg3} \cdot \frac{lg5}{lg4} \cdot \frac{lg6}{lg5} \cdot \frac{lg7}{lg6} \cdot \frac{lg8}{lg7} = \frac{1}{lg2} \cdot lg8= \frac{lg8}{lg2} = \\ \\ =log_28=log_22^3=3log_22=\boxed{3} [/tex]

Alte intrebari