(log3 2+3 log0,25):(log3 28-log3 7)

Решите задачу:

$(log_32+3log_30.25):(log_328-log_37)=...$
$(log_32+3log_3{\frac14}):log_3\frac{28}{7}=(log_32+log_3{\frac14}^3):log_3\frac{28}{7}=\\=(log_32+log_3{\frac1{64}}):log_3\frac{28}{7}=log_3{\frac2{64}:log_3\frac{28}{7}=\\=log_3{\frac1{32}:log_34=$
$=log_3{\frac1{32}:log_34=log_4\frac1{32}=\log_{2^2}\frac1{2^5}=log_{2^2}2^{-5}=-5*\frac12log_22=$
$=-\frac52=-2.5$