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Sa se calculeze:
log₃2/1+log₃3/2+log₃4/3+...........+log₃9/8
cu explicatii va rog


Răspuns :

log₃(2/1x3/2x4/3x.....x8/7x9/8)=
log₃9=2

se aplica proprietatea

[tex]\displaystyle log_3 \frac{2}{1} +log_3 \frac{3}{2} +log_3 \frac{4}{3} +...+log_3 \frac{9}{8} =\\ =log_3\left( \frac{\not2}{1} \cdot \frac{\not3}{\not2} \cdot \frac{\not4}{\not3} \cdot \frac{\not5}{\not4} \cdot \frac{\not6}{\not5} \cdot \frac{\not7}{\not6} \cdot \frac{\not8}{\not7} \cdot \frac{9}{\not8} \right) =log_39=log_33^2=2log_33=2 [/tex]