8 (125) = log0,5 (x) + log4 (0,04)...

Anna1996a @Anna1996a

July 2022 1 9 Report

log2 (4x-3) + log 1/8 (125) = log0,5 (x) + log4 (0,04)

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Ответ:

{ 1 }

Объяснение:

$\displaystyle \tt log_{2} (4 \cdot x-3)+log_{\frac{1}{8} } 125=log_{0,5} x+log_{4} 0,04$

Область допустимых значений:

$\displaystyle \tt \left \{ {{4 \cdot x-3>0} \atop {x>0}} \right. \Leftrightarrow \left \{ {{x>\frac{3}{4} } \atop {x>0}} \right. \Rightarrow x>\frac{3}{4} \Rightarrow x \in (\frac{3}{4}; +\infty).$

Решение.

$\displaystyle \tt log_{2} (4 \cdot x-3)+log_{\frac{1}{8} } 125=log_{0,5} x+log_{4} 0,04\\\\log_{2} (4 \cdot x-3)+log_{2^{-3}} 5^3=log_{2^{-1}} x+log_{2^2} 5^{-2}\\\\log_{2} (4 \cdot x-3)+\frac{3}{-3} \cdot log_{2} 5=\frac{1}{-1} \cdot log_{2} x+\frac{-2}{2} \cdot log_{2} 0,2\\\\log_{2} (4 \cdot x-3)- log_{2} 5=- log_{2} x-log_{2} 5$

$\displaystyle \tt log_{2} (4 \cdot x-3)+ log_{2} x=0\\\\log_{2} (4 \cdot x-3) \cdot x=log_{2} 1 \\\\(4 \cdot x-3) \cdot x=1\\\\4 \cdot x^2-3 \cdot x-1=0$

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.

3 votes Thanks 5

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