Himpunan penyelesaian dari pertidaksamaan 5 1 ​ lo g ( x 2 − 2 x + 1 ) ≥ 5 1 ​ lo g ( x + 11 ) adalah ....

Pembahasan

Diketahui pertidaksamaan . Dapat diperhatikan bahwa basisnya sama, yaitu . Karena berada di antara 0 dan 1, maka berlaku pertidaksamaan berikut. Oleh karena itu, didapat pertidaksamaan sebagai berikut. Didapat bahwa pembuat nolnya adalah x = − 2 atau x = 5. Kemudian, perhatikan garis bilangan berikut! Karena tanda pertidaksamannya ≤ , maka pilih daerah yang negatif, yaitu − 2 ≤ x ≤ 5. Selanjutnya, perhatikan syarat numerus dari logaritma! Numerus ruas kiri x 2 − 2 x + 1 ( x − 1 ) 2 ​ > > ​ 0 0 ​ Perhatikan bahwa pertidaksamaan di atas akan terpenuhi untuk setiap x < 1 atau x > 1. Numerus ruas kanan Kemudian, akan dicari irisan dari penyelesaian pertidaksamaan dan syarat-syarat tersebut. }{\color[rgb]{0.0, 0.5, 0.0}1}\\{\color[rgb]{0.0, 0.0, 1.0}3}{\color[rgb]{0.0, 0.0, 1.0}.}{\color[rgb]{0.0, 0.0, 1.0}\;}{\color[rgb]{0.0, 0.0, 1.0}x}{\color[rgb]{0.0, 0.0, 1.0}>}{\color[rgb]{0.0, 0.0, 1.0}-}{\color[rgb]{0.0, 0.0, 1.0}11}" data-mathml="«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mn mathcolor=¨#FF0000¨»1«/mn»«mo mathcolor=¨#FF0000¨».«/mo»«mo mathcolor=¨#FF0000¨»§#160;«/mo»«mo mathcolor=¨#FF0000¨»-«/mo»«mn mathcolor=¨#FF0000¨»2«/mn»«mo mathcolor=¨#FF0000¨»§#8804;«/mo»«mi mathcolor=¨#FF0000¨»x«/mi»«mo mathcolor=¨#FF0000¨»§#8804;«/mo»«mn mathcolor=¨#FF0000¨»5«/mn»«mspace linebreak=¨newline¨/»«mn mathcolor=¨#007F00¨»2«/mn»«mo mathcolor=¨#007F00¨».«/mo»«mo mathcolor=¨#007F00¨»§#160;«/mo»«mi mathcolor=¨#007F00¨»x«/mi»«mo mathcolor=¨#007F00¨»§#60;«/mo»«mn mathcolor=¨#007F00¨»1«/mn»«mo mathcolor=¨#007F00¨»§#160;«/mo»«mi mathcolor=¨#007F00¨»atau«/mi»«mo mathcolor=¨#007F00¨»§#160;«/mo»«mi mathcolor=¨#007F00¨»x«/mi»«mo mathcolor=¨#007F00¨»§#62;«/mo»«mn mathcolor=¨#007F00¨»1«/mn»«mspace linebreak=¨newline¨/»«mn mathcolor=¨#0000FF¨»3«/mn»«mo mathcolor=¨#0000FF¨».«/mo»«mo mathcolor=¨#0000FF¨»§#160;«/mo»«mi mathcolor=¨#0000FF¨»x«/mi»«mo mathcolor=¨#0000FF¨»§#62;«/mo»«mo mathcolor=¨#0000FF¨»-«/mo»«mn mathcolor=¨#0000FF¨»11«/mn»«/math»" role="math" src="data:image/png;base64,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" style="max-width: none;"> Perhatikan garis bilangan berikut! Berdasarkan gambar di atas, dapat dilihat bahwa ketiga syarat tersebut terpenuhi saat − 2 ≤ x < 1 atau 1 < x ≤ 5. Dengan demikian, himpunan penyelesaian dari pertidaksamaan di atas adalah { x ∣ − 2 ≤ x < 1 atau 1 < x ≤ 5 , x ∈ R } . Jadi, jawaban yang tepat adalah D.