Log 0,5 (x-1)=2 log x+2 (16)=4 log 3x-1 (4)=2 lg(2-x)+lg(1-x)=lg(12) log 4 (40+8log...

Question 1

Log 0,5 (x-1)=2
log x+2 (16)=4
log 3x-1 (4)=2
lg(2-x)+lg(1-x)=lg(12)
log 4 (40+8log 3(x+4))=3
log 3(log 2(x-4))=0
Пожалуйста помогите
Напишите с решением)

Question 2

Решите задачу:

$1)\; \; log_{0,5}(x-1)=2\; ,\; \; \; ODZ:\; x\ \textgreater \ 1\\\\x-1=(0,5)^2\; ,\; \; x-1=0,25\; ,\; \; x=1,25\\\\2)\; \; log_{x+2}16=4\; ,\; \; \; ODZ:\; \left \{ {{x+2\ \textgreater \ 0} \atop {x+2\ne 1}} \right. \; \left \{ {{x\ \textgreater \ -2} \atop {x\ne -1}} \right. \\\\(x+2)^4=16\; ,\; \; (x+2)^4=2^4\; ,\; \; x+2=\pm 2\\\\x_1=0\; ,\; \; x_2=-4\; \notin ODZ\\\\Otvet:\; x=0\; .\\\\3)\; \; log\, _{3x-1}\; 4=2\; ,\; \; \; ODZ:\; \left \{ {{3x-1\ \textgreater \ 0} \atop {3x-1\ne 1}} \right. \; \left \{ {{x\ \textgreater \ \frac{1}{3}} \atop {x\ne \frac{2}{3}}} \right.$

$(3x-1)^2=4\; ,\; \; (3x-1)^2-2^2=0\\\\(3x-1-2)(3x-1+2)=0\\\\(3x-3)(3x+1)=0\\\\x_1=1\; ,\; x_2=-\frac{1}{3}\notin ODZ\\\\Otvet:\; x=1\; .$

$4)\; \; lg(2-x)+lg(1-x)=lg12\; ,\; \; ODZ: \left \{ {{2-x\ \textgreater \ 0} \atop {1-x\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textless \ 2} \atop {x\ \textless \ 1}} \right. \; \to \; x\ \textless \ 1\\\\lg(2-x)(1-x)=lg12\\\\(2-x)(1-x)=12\\\\2-3x+x^2-12=0\\\\x^2-3x-10=0\\\\x_1=-2\; ,\; \; x_2=5\notin ODZ\\\\Otvet:\; \; x=-2\; .$

$5)\; \; log_4(40+8log_3(x+4))=3\; ,\\\\ODZ:\; \left \{ {{x+4\ \textgreater \ 0} \atop {40+8log_3(x+4)\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ -4} \atop {log_3(x+4)\ \textgreater \ -5}} \right. \; \left \{ {{x\ \textgreater \ -4} \atop {x+4\ \textgreater \ 3^{-5}}} \right. \; \left \{ {{x\ \textgreater \ -4} \atop {x\ \textgreater \ -3\frac{242}{243}}} \right. \\\\x\ \textgreater \ -3\frac{242}{243}\\\\40+8log_3(x+4)=4^3\\\\8log_3(x+4)=64-40\\\\log_3(x+4)=3\\\\x+4=3^3\\\\x=27-4\\\\\underline {x=23}\in ODZ$

$6)\; \; log_3(log_2(x-4))=0\; ,\; \; ODZ:\; \left \{ {{x-4\ \textgreater \ 0} \atop {log_2(x-4)\ \textgreater \ 0}} \right. \; \left \{ {{x\ \textgreater \ 4} \atop {x-4\ \textgreater \ 1}} \right. \; \to \; x\ \textgreater \ 5\\\\log_2(x-4)=1\\\\x-4=2\\\\\underline {x=6}\in ODZ$

Question 3

Спасибо