308 полностью= 1 пример там309-не надо 310-полностью

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308)
$\log_{49}28=\log_{7^2}28= \frac{1}{2} \log_{7}(4\cdot7)= \frac{1}{2}( \log_{7}(4)+\log_{7}(7))
\\\
= \frac{1}{2}( \log_{7}2^2+1)= \frac{1}{2}( 2\log_{7}2+1) =\frac{1}{2}( 2m+1)=m+ \frac{1}{2}$

310)
$\log_62=\frac{1}{\log_26}= \frac{1}{\log_2(3\cdot2)}=\frac{1}{\log_23+\log_22}=\frac{1}{\log_23+1}=m
\\\
\Rightarrow\log_23=\frac{1}{m}-1=\frac{1-m}{m}
\\\\
\log_{24}72=\log_{24}(24\cdot3)=\log_{24}24+\log_{24}3=1+ \frac{1}{\log_324}=
\\\
=1+ \frac{1}{\log_{3}(3\cdot8)}=1+\frac{1}{\log_33+\log_38}=1+\frac{1}{1+\log_32^3}=
\\\
=1+\frac{1}{1+3\log_32}=1+\cfrac{1}{1+\frac{3}{\log_23} }$

$\log_{24}72=1+\cfrac{1}{1+\cfrac{3}{\frac{1-m}{m}}}=1+\cfrac{1}{1+\frac{3m}{1-m}}=1+\cfrac{1}{\frac{1-m+3m}{1-m}}=
\\\\
=1+\frac{1-m}{1+2m}=\frac{1+2m+1-m}{1+2m}=\frac{2+m}{1+2m}$

312)
$\frac{\log_2192}{\log_{12}2} - \frac{\log_224}{\log_{96}2} =
\log_2192\log_212 - \log_224\log_296 =
\\\
=\log_2(3\cdot2^6)\log_2(3\cdot2^2) - \log_2(3\cdot2^3)\log_2(3\cdot2^5) =
\\\
=(\log_23+\log_22^6)(\log_23+\log_22^2) - (\log_23+\log_22^3)(\log_23+\log_22^5) 
\\\
=(\log_23+6)(\log_23+2) - (\log_23+3)(\log_23+5) =
\\\
=\log_2^23+2\log_23+6\log_23+12-\log_2^23-5\log_23-3\log_23-15=-3$

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