Question

Помогите, пожалуйста, с алгеброй.

Answer 1

Формулы:
$log_abc=log_ab+log_ac \log_a frac{b}{c} =log_ab-log_ac \ log_ab^c=clog_ab \ log_aa=1$

$log_327 sqrt{a^2b} =log_327a sqrt{b} =log_327+log_3a +log_3 sqrt{b} = \ =log_33^3+log_3a +log_3 b^ frac{1}{2} =3+log_3a + frac{1}{2}log_3 b$

$log_3 frac{ sqrt[3]{a} }{9b^2} =log_3 sqrt[3]{a} -log_39b^2= log_3a^ frac{1}{3} -(log_39+log_3b^2)= \ =frac{1}{3}log_3a -(2+2log_3b)=frac{1}{3}log_3a -2log_3b-2$

$log_3 frac{81 sqrt{a^3} }{ sqrt[3]{b^2}} =log_381 sqrt{a^3}-log_3 sqrt[3]{b^2}= (log_381 +log_3sqrt{a^3})-log_3 b^ frac{2}{3} = \ =log_33^4 +log_3a^ frac{3}{2} -log_3 b^ frac{2}{3} = 4 + frac{3}{2} log_3a-frac{2}{3}log_3 b$

$log_3 frac{ sqrt[4]{ab^2} }{ sqrt[3]{a^5b} } =log_3sqrt[4]{ab^2}-log_3 sqrt[3]{a^5b}= \ =(log_3sqrt[4]{a}+log_3 sqrt[4]{b^2} )-(log_3 sqrt[3]{a^5}+log_3 sqrt[3]{b} )= \ =log_3a^ frac{1}{4} +log_3b^ frac{1}{2}-log_3 a^ frac{5}{3}-log_3 b^ frac{1}{3} = \ = frac{1}{4} log_3a+ frac{1}{2}log_3b-frac{5}{3}log_3 a-frac{1}{3}log_3 b= \ =( frac{1}{4}- frac{5}{3} ) log_3a+ (frac{1}{2}- frac{1}{3} )log_3b=$
$= frac{3-20}{12} log_3a+frac{3-2}{6}log_3b=- frac{17}{12} log_3a+frac{1}{6}log_3b$