[tex]f(x)=log_2(x-4)^2-log_2(x-5)=log_2\frac{(x-4)^2}{x-5}[/tex]
dziedzina
x ≠ 4
x>5
[tex]x\in(5;+\infty)[/tex]
[tex]f(x)-log_2m=0\\\\m>0\\\\log_2\frac{(x-4)^2}{x-5}-log_2m=0\\\\log_2\frac{(x-4)^2}{m(x-5)}=log_21\\\\\frac{(x-4)^2}{m(x-5)}=1\\\\(x-4)^2=m(x-5)\\\\x^2-8x+16=mx-5m\\\\x^2-8x+16-mx+5m=0\\\\x^2-(8+m)x+16+5m=0\\\\\Delta=[-(8+m)]^2-4\cdot1\cdot(16+5m)=64+16m+m^2-64-20m=m^2-4m\\\\m(m-4)>0\\\\m\in(-\infty;0)\cup(4;+\infty)[/tex]
Po uwzględnieniu m>0
[tex]m\in(4;+\infty)[/tex]
