Tentukan batas-batas nilai x yang memenuhi pertidaksamaanberikut. lo g ( 3 + x ) + lo g 4 ≥ 2 lo g x

Pembahasan

Ingat bahwa! a lo g b + a lo g c = a lo g ( b × c ) Untuk a > 1 JIka a lo g f ( x ) ≥ a lo g g ( x ) ⇔ f ( x ) ≥ g ( x ) , dengan f ( x ) > 0 dan g ( x ) > 0 Sehingga nilai x dapat ditentukan dengan cara berikut. syarat numerusnya. &0\\4x&>&-12\\x&>&-{3.......{\color[rgb]{0.0, 0.0, 1.0}2}{\color[rgb]{0.0, 0.0, 1.0})}}\end{array}" data-mathml="«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mn»12«/mn»«mo»+«/mo»«mn»4«/mn»«mi»x«/mi»«/mtd»«mtd»«mo»§#62;«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mn»4«/mn»«mi»x«/mi»«/mtd»«mtd»«mo»§#62;«/mo»«/mtd»«mtd»«mo»-«/mo»«mn»12«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»x«/mi»«/mtd»«mtd»«mo»§#62;«/mo»«/mtd»«mtd»«mo»-«/mo»«mrow»«mn»3«/mn»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mn mathcolor=¨#0000FF¨»2«/mn»«mo mathcolor=¨#0000FF¨»)«/mo»«/mrow»«/mtd»«/mtr»«/mtable»«/math»" role="math" src="data:image/png;base64,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" style="max-width: none;"> dan &0\\x&>&0\;......{\color[rgb]{0.0, 0.0, 1.0}3}{\color[rgb]{0.0, 0.0, 1.0})}\end{array}" data-mathml="«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msup»«mi»x«/mi»«mn»2«/mn»«/msup»«/mtd»«mtd»«mo»§#62;«/mo»«/mtd»«mtd»«mn»0«/mn»«/mtd»«/mtr»«mtr»«mtd»«mi»x«/mi»«/mtd»«mtd»«mo»§#62;«/mo»«/mtd»«mtd»«mn»0«/mn»«mo»§#160;«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mo».«/mo»«mn mathcolor=¨#0000FF¨»3«/mn»«mo mathcolor=¨#0000FF¨»)«/mo»«/mtd»«/mtr»«/mtable»«/math»" role="math" src="data:image/png;base64,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" style="max-width: none;"> Dari 1, 2, dan 3 diperoleh gambar himpunan penyelesaian sebagai berikut. { x ∣0 < x ≤ 6 , x ∈ R } Dengan demikian himpunan penyelesaiandari pertidaksamaanlogaritma tersebut adalah { x ∣0 < x ≤ 6 , x ∈ R }