Question

Salut! Stiind ca logaritm de baza3 din2 =a, aratati ca logaritm baza 3 din 8 + logaritm baza3 din 100 - logaritm baza 3 din 25 =5a (Va rog, sa-mi rezolvati mai detaliat) CORONITA

Answer 1

$=\log_38+\log_3100+\log_325$
$=\log_32^3+\log_310^2-\log_35^2$
$=3\log_32+2\log_310-2\log_35$
$=3a+2\left ( \log_310-2\log_35 \right )$
$=3a+2\left ( \log_3\frac{10}{5} \right )$
$=3a+2\log_32$
$=3a+2a$
$=5a$

Answer 2

$\displaystyle log_32=a \\ log_38+log_3100-log_325=5a \\ --------------- \\ log_38+log_3100-log_325=log_3(8 \cdot 100)-log_325=log_3800-log_325= \\ =log_3 \frac{800}{25} =log_332=log_3 2^5=5log_32=5a$