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Question 1

Question 2

Решите задачу:

$1)\; \; 10^{-log_{10}0,5}\cdot log_54+log_50,5=10^{log_{10}(\frac{1}{2})^{-1}}\cdot log_52^2+log_5\frac{1}{2}=\\\\=(\frac{1}{2})^{-1}\cdot 2log_52+log_52^{-1}=2\cdot 2\cdot log_52-log_52=3\cdot log_52\; ;\\\\\\log_56-log_512=log_5(3\cdot 2)-log_5(3\cdot 2^2)=\\\\=log_53+log_52-(log_53+2log_52)=-log_52\; ;\\\\\\ \frac{10^{-log_{10}0,5}\cdot log_54+log_50,5}{log_56-log_512} = \frac{3\cdot log_52}{-log_52} =-3\; ;$

$2)\; \; log_{0,5}3-log_{0,5}21=log_{1/2}3-log_{1/2}(3\cdot 7)=\\\\=log_{2^{-1}}3-log_{2^{-1}}(3\cdot 7)=-log_23+log_2(3\cdot 7)=\\\\=-log_23+log_23+log_27=log_27\\\\\\log_{0,5}1,4+10^{2log_{10}0,5}\cdot log_{0,5}625=-log_2\frac{14}{10}+10^{log_{10}(0,5)^2}\cdot (-log_25^4)=\\\\=-\Big (log_2(2\cdot 7)-log_2(2\cdot 5)\Big )+(0,5)^2\cdot (-4log_25)=\\\\=-log_22-log_27+log_22+log_25-0,25\cdot 4log_25=\\\\=-log_27+log_25-log_25=-log_27$

$\frac{log_{0,5}3-log_{0,5}21}{log_{0,5}1,4+10^{2log_{10}0,5}\cdot log_{0,5}625} =\frac{log_27}{-log_27}=-1$

$3)\; \; \frac{log_4log_481}{1+log_4log_43}= \frac{log_4(log_43^4)}{1+log_4(log_{2^2}3)} = \frac{log_4(4log_43)}{1+log_4(\frac{1}{2}log_23)} =\\\\= \frac{log_4(4log_{2^2}3)}{1+log_{2^2}2^{-1}+log_{2^2}(log_23)} = \frac{log_{2^2}(2log_23)}{1-\frac{1}{2}+\frac{1}{2}log_2(log_23)} = \frac{\frac{1}{2}(log_22+log_2(log_23))}{\frac{1}{2}(1+log_2(log_23))} =\\\\= \frac{1+log_2(log_23)}{1+log_2(log_23)} =1$

$4)\; \; \frac{log_3log_3125}{1+log_3log_35}=\frac{log_3(log_35^3)}{1+log_3(log_35)}=\frac{log_3(3\cdot log_35)}{1+log_3(log_35)}=\\\\=\frac{log_33+log_3(log_35)}{1+log_3(log_35)}=\frac{1+log_3(log_35)}{1+log_3(log_35)}=1\\\\P.S.\; 3)\; \; \frac{log_4log_481}{1+log_4log_43} = \frac{log_4(log_43^4)}{1+log_4(log_43)} = \frac{log_4(4\cdot log_43)}{1+log_4(log_43)} =\\\\= \frac{\overbrace{log_44}^{1}+log_4(log_43)}{1+log_4(log_43)} =1$

Question 3

ты сидишь вк?)

Question 4

Во 2 пункте описка в условии

Question 5

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