Question

решить логарифмическое выражение​

Answer 1

Ответ:

$-3$

Объяснение:

$\frac{log_2192}{log_{12}2}-\frac{log_224}{log_{96}2}=\frac{log_2192}{\frac{1}{log_212}}-\frac{log_224}{\frac{1}{log_296}}=log_2192*log_212-log_224*log_296=\\\\=log_2(3*2^6)*log_2(3*2^2)-log_2(3*2^3)*log_2(3*2^5)=\\\\=(log_23+6log_22)(log_23+2log_22)-(log_23+3log_22)(log_23+5log_22)=\\\\=(log_23+6)(log_23+2)-(log_23+3)(log_23+5)=\\\\=(log_2^23+8log_23+12)-(log_2^23+8log_23+15)=\\\\=log_2^23+8log_23+12-log_2^23-8log_23-15=-3$

Answer 2

$\frac{log_{2}192 }{log_{12}2}-\frac{log_{2}24}{log_{96}2}=log_{2}192*log_{2}12-log_{2}24*log_{2}96=log_{2}(2^{5}*6)*log_{2}(2*6)-log_{2}(2^{2}*6)*log_{2}(2^{4}*6)=(log_{2}2^{5}+log_{2} 6)(log_{2} 2+log_{2}6)-(log_{2}2^{2}+log_{2}6)(log_{2}2^{4} +log_{2}6)=(5+log_{2}6)(1+log_{2}6)-(2+log_{2} 6)(4+log_{2}6)=5+5log_{2}6+log_{2}6+log^{2}_{2}6-8-2log_{2}6-4log_{2}6-log^{2}_{2}6=-3$