Question

aratati ca logaritm in baza 3 din1/9 +radical de ordin trei din 27 =1

Answer 1

$log_3 \frac{1}{9}+ \sqrt[3]{27}=1\ \textless \ =\ \textgreater \ -2+3=1 \ \textless \ =\ \textgreater \ 1=1 (A) ~log_3 \frac{1}{9}=log_3 (\frac{1}{3}) ^2$
[tex]2log_3 \frac{1}{3}=2*(-1) ~ ~ \sqrt[3]{27}= \sqrt[3]{3^3}=3^ \frac{3}{3} =3 [/tex]

Answer 2

$\it log_3 \dfrac{1}{9} +\sqrt[3]{27} = log_31 - log_39+3 = 0-2+3=1$