log
2
(4⋅x−3)+log
8
1
125=log
0,5
x+log
4
0,04
Область допустимых значений:
\displaystyle \tt \left \{ {{4 \cdot x-3 > 0} \atop {x > 0}} \right. \Leftrightarrow \left \{ {{x > \frac{3}{4} } \atop {x > 0}} \right. \Rightarrow x > \frac{3}{4} \Rightarrow x \in (\frac{3}{4}; +\infty).{
x>0
4⋅x−3>0
⇔{
x>0
x>
4
3
⇒x>
4
3
⇒x∈(
4
3
;+∞).
Решение.
\begin{gathered}\displaystyle \tt log_{2} (4 \cdot x-3)+log_{\frac{1}{8} } 125=log_{0,5} x+log_{4} 0,04log_{2} (4 \cdot x-3)+log_{2^{-3}} 5^3=log_{2^{-1}} x+log_{2^2} 5^{-2}log_{2} (4 \cdot x-3)+\frac{3}{-3} \cdot log_{2} 5=\frac{1}{-1} \cdot log_{2} x+\frac{-2}{2} \cdot log_{2} 0,2log_{2} (4 \cdot x-3)- log_{2} 5=- log_{2} x-log_{2} 5\end{gathered}
log
2
(4⋅x−3)+log
8
1
125=log
0,5
x+log
4
0,04
log
2
(4⋅x−3)+log
2
−3
5
3
=log
2
−1
x+log
2
2
5
−2
log
2
(4⋅x−3)+
−3
3
⋅log
2
5=
−1
1
⋅log
2
x+
2
−2
⋅log
2
0,2
log
2
(4⋅x−3)−log
2
5=−log
2
x−log
2
5
\begin{gathered}\displaystyle \tt log_{2} (4 \cdot x-3)+ log_{2} x=0log_{2} (4 \cdot x-3) \cdot x=log_{2} 1 (4 \cdot x-3) \cdot x=14 \cdot x^2-3 \cdot x-1=0\end{gathered}
log
2
(4⋅x−3)+log
2
x=0
log
2
(4⋅x−3)⋅x=log
2
1
(4⋅x−3)⋅x=1
4⋅x
2
−3⋅x−1=0
D=(-3)²-4·4·(-1)=9+16=25=5²
x₁=(3-5)/(2·4)=-2/8= -1/4 ∉ ОДЗ,
x₂=(3+5)/(2·4)=8/8= 1 ∈ ОДЗ.
