Срочно!1-3-4 задания

Решите задачу:

$1.\;\log_27-\log_263+2\log_26=\log_27-\log_263+\log_236=\log_2\left(\frac{7\cdot36}{63}\right)=\\=\log_24=\log_22^2=2\\\\3.a)\;5^x-7\cdot5^{x-2}=90\\5^x-7\cdot5^x\cdot5^{-2}=90\\5^x-\frac7{25}\cdot5^x=90\\\frac{18}{25}\cdot5^x=90\\5^x=90:\frac{18}{25}\\5^x=90\cdot\frac{25}{18}\\5^x=5\cdot25\\5^x=125\\5^x=5^3\\x=3$

$b)\;\log_\frac12(x^2-4x-1)=-2\\O.D.3.:\\x^2-4x-1\ \textgreater \ 0\\(x-2-\sqrt5)(x-2+\sqrt5)\ \textgreater \ 0\\x\in(-\infty;2-\sqrt5)\cup(2+\sqrt5;\;+\infty)\\\\\log_{2^{-1}}(x^2-4x-1)=-2\\-\log_2(x^2-4x-1)=-2\\\log_2(x^2-4x-1)=2\\x^2-4x-1=4\\x^2-4x-5=0\\D=16 +20=36\\x_{1,2}=\frac{4\pm6}2\\x_1=-1\\x_2=5$

$4.\;a)\;9^{0,5x^2-3}\ \textless \ 27\\3^{2(0,5x^2-3)}\ \textless \ 3^3\\x^2-6\ \textless \ 3\\x^2\ \textless \ 9\\-3\ \textless \ x\ \textless \ 3\\x\in(-3;\;3)\\\\b)\;\log_\frac17(3-x)\ \textgreater \ -1\\\log_{7^{-1}}(3-x)\ \textgreater \ -1\\-\log_7(3-x)\ \textgreater \ -1\\\log_7(3-x)\ \textless \ 1\\3-x\ \textless \ 7\\x\ \textgreater \ -4$