Jawab:
[tex]\large\text{$\begin{aligned}&x<-\tfrac{195}{7}\\&\textsf{Notasi interval:}\ \left({-}\infty,\: {-}\tfrac{195}{7}\right)\end{aligned}$}[/tex]
Penjelasan dengan langkah-langkah:
[tex]\large\text{$\begin{aligned}&\log\left[\tfrac{1}{2}(5-7x)\right]>2\\&\log\left[\tfrac{1}{2}(5-7x)\right]>\log100\\&\tfrac{1}{2}(5-7x)>100\\&5-7x>200\\&{-}7x>195\\&x<\bf-\frac{195}{7}\qquad(\textsf{kedua ruas dibagi $-7$})\end{aligned}$}[/tex]
ATAU
[tex]\large\text{$\begin{aligned}&\log\left[\tfrac{1}{2}(5-7x)\right]>2\\&\log\tfrac{1}{2}+\log(5-7x)>\log100\\&\log2^{-1}+\log(5-7x)>\log100\\&{-}\log2+\log(5-7x)>\log100\\&\log(5-7x)>\log100+\log2\\&\log(5-7x)>\log(100\cdot2)\\&\log(5-7x)>\log200\\&5-7x>200\\&{-}7x>195\\&x<\bf-\frac{195}{7}\qquad(\textsf{kedua ruas dibagi $-7$})\end{aligned}$}[/tex]
∴ Maka, batas-batas nilai x yang memnuhi perhitungan logaritma dari log½(5-7x) > 2 adalah:
[tex]\large\text{$\begin{aligned}&x<-\tfrac{195}{7}\\&\textsf{Notasi interval:}\ \left({-}\infty,\: {-}\tfrac{195}{7}\right)\end{aligned}$}[/tex]
[answer.2.content]
