Question

Пожалуйста решите уравнение log2 (x^2+8)-log2 (x-1)=log0,5 1/8

Answer 1

Log2 (x^2+8)-log2 (x-1)=log0,5 1/8
ОДЗ:
$\left \{ {{x^2+8\ \textgreater \ 0} \atop {x-1\ \textgreater \ 0}} \right. = \left \{ {{x^2+8\ \textgreater \ 0} \atop {x\ \textgreater \ 1}} \right$)
x∈(-1;+∞)
log2((x^2+8)/(x-1))=-log2(1/8)
((x^2+8)/(x-1))-8=0
(x^2+8-8x+8)/(x-1)=0
(x^2-8x+16)/(x-1)=0
(x-4)^2/(x-1)=0
x=4(x∈ОДЗ)
Ответ: х=4