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log in baza 2 din 12 + log in baza 2 din 14 - log in baza 2 din 21 = 3 ..ajutor !

Răspuns :

[tex]\log_2 12 + \log_2 14- \log_2 21= \log_2 12 \cdot14:21 =\log_2 8 \\ \\ \boxed{\log_28= 3} \\ \\ \text{S-a demonstrat ca \log_28=3} \\ \\ [/tex]
[tex]\displaystyle log_212+log_214-log_221=log_2(12 \cdot 14)-log_221= \\ =log_2168-log_221=log_2 \frac{168}{21} =log_28=log_22^3=3log_22=3 [/tex]