Tentukan penyelesaian persamaan berikut. b.

Pertanyaan

Tentukan penyelesaian persamaan berikut.

b. $log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus log presuperscript begin inline style 1 half end style end presuperscript space open parentheses x plus 10 close parentheses equals 0$

Pembahasan Soal:

Ingat sifat bentuk logaritma:

$table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript a space b times log presuperscript a space c end cell equals cell log presuperscript a space b c end cell row cell log presuperscript a to the power of n end presuperscript space b to the power of m end cell equals cell m over n space log presuperscript a space b end cell end table$

Ingat ada persamaan logaritma berlaku jika $log presuperscript a space f open parentheses x close parentheses equals log presuperscript a space g open parentheses x close parentheses comma space a greater than 0$ dan $a not equal to 1$ maka $f open parentheses x close parentheses equals g open parentheses x close parentheses$ dengan syarat $f open parentheses x close parentheses greater than 0$ dan $g open parentheses x close parentheses greater than 0$ .

Diketahui $log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus log presuperscript begin inline style 1 half end style end presuperscript space open parentheses x plus 10 close parentheses equals 0$ maka:

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus log presuperscript begin inline style 1 half end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus log presuperscript begin inline style 2 to the power of negative 1 end exponent end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus open parentheses fraction numerator 1 over denominator negative 1 end fraction close parentheses log presuperscript begin inline style 2 end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses minus log presuperscript begin inline style 2 end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses end cell equals cell log presuperscript begin inline style 2 end style end presuperscript space open parentheses x plus 10 close parentheses end cell row cell log presuperscript 2 space open parentheses x plus 7 close parentheses open parentheses x plus 6 close parentheses end cell equals cell log presuperscript 2 space open parentheses x plus 10 close parentheses end cell row cell log presuperscript 2 space open parentheses x squared plus 13 x plus 42 close parentheses end cell equals cell log presuperscript 2 space open parentheses x plus 10 close parentheses end cell row cell x squared plus 13 x plus 42 end cell equals cell x plus 10 end cell row cell x squared plus 13 x minus x plus 42 minus 10 end cell equals 0 row cell x squared plus 12 x plus 32 end cell equals 0 row cell open parentheses x plus 8 close parentheses open parentheses x plus 4 close parentheses end cell equals 0 end table end style$

diperoleh nilai yang memenuhi $x equals negative 8$ atau $x equals negative 4$

Syarat numerus 1)

$table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 7 end cell greater than 0 row x greater than cell negative 7 end cell end table$

Syarat numerus 3)

$table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 6 end cell greater than 0 row x greater than cell negative 6 end cell end table$

Syarat numerus 2)

$table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 10 end cell greater than 0 row x greater than cell negative 10 end cell end table$

Dari syarat numerus mengharuskan $x greater than negative 6$ maka nilai yang memenuhi adalah $x equals negative 4$ .

Dengan demikian himpunan penyelesaiany adalah $open curly brackets negative 4 close curly brackets$ .

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

S. Amamah

Mahasiswa/Alumni Universitas Negeri Malang

Terakhir diupdate 11 Juli 2021

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Jika , tentukan nilai yang memenuhi persamaan tersebut.

Pembahasan Soal:

Ingat sifat-sifat logaritma:

$table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript a space b minus log presuperscript a space c end cell equals cell log presuperscript a space b over c end cell row cell log presuperscript a space a to the power of n end cell equals n row cell fraction numerator 1 over denominator log presuperscript a space b end fraction end cell equals cell log presuperscript b space a end cell row cell log presuperscript a to the power of m end presuperscript space b to the power of n end cell equals cell n over m log presuperscript a space b end cell end table$

Ingat ada persamaan logaritma berlaku jika $log presuperscript h open parentheses x close parentheses end presuperscript space f open parentheses x close parentheses equals log presuperscript h open parentheses x close parentheses end presuperscript space g open parentheses x close parentheses comma space a greater than 0$ dan $a not equal to 1$ maka $f open parentheses x close parentheses equals g open parentheses x close parentheses$ dengan syarat $h open parentheses x close parentheses greater than 0 comma space h open parentheses x close parentheses not equal to 1$ , $f open parentheses x close parentheses greater than 0$ dan $g open parentheses x close parentheses greater than 0$ .

Diketahui $fraction numerator 1 over denominator log presuperscript 16 space open parentheses x minus 2 close parentheses minus log presuperscript 16 space open parentheses x squared minus 4 x plus 4 close parentheses end fraction equals negative 2$ maka:

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator log presuperscript 16 space open parentheses x minus 2 close parentheses minus log presuperscript 16 open parentheses x squared minus 4 x plus 4 close parentheses end fraction end cell equals cell negative 2 end cell row cell fraction numerator 1 over denominator log presuperscript 16 space begin display style fraction numerator x minus 2 over denominator x squared minus 4 x plus 4 end fraction end style end fraction end cell equals cell negative 2 end cell row cell fraction numerator 1 over denominator log presuperscript 16 space begin display style fraction numerator x minus 2 over denominator open parentheses x minus 2 close parentheses open parentheses x minus 2 close parentheses end fraction end style end fraction end cell equals cell negative 2 end cell row cell fraction numerator 1 over denominator log presuperscript 16 space begin display style fraction numerator 1 over denominator x minus 2 end fraction end style end fraction end cell equals cell negative 2 end cell row cell fraction numerator 1 over denominator log presuperscript 16 space open parentheses x minus 2 close parentheses to the power of negative 1 end exponent end fraction end cell equals cell negative 2 end cell row cell log presuperscript open parentheses x minus 2 close parentheses to the power of negative 1 end exponent end presuperscript space 16 end cell equals cell negative 2 end cell row cell negative log presuperscript open parentheses x minus 2 close parentheses end presuperscript space 16 end cell equals cell negative 2 end cell row cell log presuperscript open parentheses x minus 2 close parentheses end presuperscript space 16 space end cell equals 2 row cell log presuperscript open parentheses x minus 2 close parentheses end presuperscript space 16 end cell equals cell log presuperscript open parentheses x minus 2 close parentheses end presuperscript space open parentheses x minus 2 close parentheses squared end cell row 16 equals cell open parentheses x minus 2 close parentheses squared end cell row 16 equals cell x squared minus 4 x plus 4 end cell row cell x squared minus 4 x plus 4 minus 16 end cell equals 0 row cell x squared minus 4 x minus 12 end cell equals 0 row cell open parentheses x minus 6 close parentheses open parentheses x plus 2 close parentheses end cell equals 0 end table$

diperoleh nilai yang memenuhi $x equals 6$ atau $x equals negative 2$

Syarat numerus 1)

$table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell greater than 0 row x greater than 2 end table$

syarat numerus 2)

$table attributes columnalign right center left columnspacing 0px end attributes row cell x squared minus 4 x plus 4 end cell greater than 0 row cell open parentheses x minus 2 close parentheses open parentheses x minus 2 close parentheses end cell greater than 0 row x greater than 2 end table$

Dari syarat numerus mengharuskan $x greater than 2$ . Dengan demikian nilai $x$ yang memenuhi adalah $x equals 6$ .

Roboguru

Akar-akar dari adalah dan maka

Pembahasan Soal:

Ingat sifat pada persamaan bentuk logaritma yaitu $scriptbase log invisible function application f left parenthesis x right parenthesis end scriptbase presuperscript a equals scriptbase log invisible function application g left parenthesis x right parenthesis end scriptbase presuperscript b rightwards arrow f open parentheses x close parentheses equals g left parenthesis x right parenthesis$ dan sifat pada bentuk logaritma yaitu

$scriptbase log invisible function application a end scriptbase presuperscript a equals 1$
$scriptbase log invisible function application b end scriptbase presuperscript a plus scriptbase log invisible function application c end scriptbase presuperscript a equals scriptbase log invisible function application b c end scriptbase presuperscript a$
$scriptbase log invisible function application b to the power of m end scriptbase presuperscript a to the power of n end presuperscript equals m over n times scriptbase log invisible function application b end scriptbase presuperscript a$
$scriptbase log invisible function application b to the power of m end scriptbase presuperscript a equals m times scriptbase log invisible function application b end scriptbase presuperscript a$

Ingat syarat basis pada logaritma $scriptbase log invisible function application b end scriptbase presuperscript a$ yaitu $a greater than 0$ dan $a not equal to 1$ dan syarat numerus pada logaritma $scriptbase log invisible function application b end scriptbase presuperscript a$ yaitu $b greater than 0$ . Sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end cell equals cell 1 plus scriptbase log invisible function application left parenthesis x minus 1 right parenthesis end scriptbase presuperscript 9 end cell row cell scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end cell equals cell scriptbase log invisible function application 3 end scriptbase presuperscript 3 plus scriptbase log invisible function application left parenthesis x minus 1 right parenthesis end scriptbase presuperscript 3 squared end presuperscript end cell row cell scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end cell equals cell scriptbase log invisible function application 3 end scriptbase presuperscript 3 plus 1 half scriptbase log invisible function application left parenthesis x minus 1 right parenthesis end scriptbase presuperscript 3 end cell row cell scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end cell equals cell scriptbase log invisible function application 3 end scriptbase presuperscript 3 plus scriptbase log invisible function application open parentheses left parenthesis x minus 1 right parenthesis to the power of 1 half end exponent close parentheses end scriptbase presuperscript 3 end cell row cell scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end cell equals cell scriptbase log invisible function application open parentheses 3 times left parenthesis x minus 1 right parenthesis to the power of 1 half end exponent close parentheses end scriptbase presuperscript 3 end cell row cell x plus 1 end cell equals cell 3 times left parenthesis x minus 1 right parenthesis to the power of 1 half end exponent end cell row cell open parentheses x plus 1 close parentheses squared end cell equals cell open parentheses 3 times left parenthesis x minus 1 right parenthesis to the power of 1 half end exponent close parentheses squared end cell row cell x squared plus 2 x plus 1 end cell equals cell 3 squared times left parenthesis x minus 1 right parenthesis end cell row cell x squared plus 2 x plus 1 end cell equals cell 9 times left parenthesis x minus 1 right parenthesis end cell row cell x squared plus 2 x plus 1 end cell equals cell 9 x minus 9 end cell row cell x squared plus 2 x minus 9 x plus 1 plus 9 end cell equals 0 row cell x squared minus 7 x plus 10 end cell equals 0 row cell left parenthesis x minus 2 right parenthesis left parenthesis x minus 5 right parenthesis end cell equals 0 row cell x minus 2 end cell equals cell 0 blank atau blank x minus 5 equals 0 end cell row x equals cell 2 blank atau blank x equals 5 end cell end table$

Didapatkan akar-akar dari persamaan tersebut yaitu $x subscript 1 equals 2$ dan $x equals 5$ . Maka nilai dari $x subscript 1 plus x subscript 2$ sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell 2 plus 5 end cell row blank equals 7 end table$

Jadi, dapat disimpulkan bahwa nilai dari $x subscript 1 plus x subscript 2$ adalah $7$ .

Oleh karena itu, jawaban yang benar adalah D.

Roboguru

Misal persamaan logaritma: mempunyai akar-akar dan maka

Pembahasan Soal:

Ingat materi persamaan logaritma dengan basisnya adalah fungsi yang berbeda dan numerusnya adalah fungsi yang sama dan bentuk umum atau definisi dari logaritma yaitu $scriptbase log invisible function application b end scriptbase presuperscript a equals c rightwards arrow a to the power of c equals b$ .

Ingat juga sifat-sifat pada bentuk logaritma yaitu

$scriptbase log invisible function application a end scriptbase presuperscript a equals 1$
$scriptbase log invisible function application b to the power of m end scriptbase presuperscript a equals m times scriptbase log invisible function application b end scriptbase presuperscript a$
$scriptbase log invisible function application b to the power of m end scriptbase presuperscript a to the power of n end presuperscript equals m over n times scriptbase log invisible function application b end scriptbase presuperscript a$

Untuk menyelesaikan persamaan tersebut maka dilakkukan permisalan terlebih dahulu.

Misalkan $scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 equals y$ sehingga

$table attributes columnalign right center left columnspacing 0px end attributes row cell scriptbase log invisible function application open parentheses x plus 1 close parentheses end scriptbase presuperscript 3 plus fraction numerator 3 over denominator scriptbase log invisible function application open parentheses x plus 1 close parentheses end scriptbase presuperscript 9 end fraction end cell equals 5 row cell scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 plus fraction numerator 3 over denominator scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 squared end presuperscript end fraction end cell equals 5 row cell scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 plus fraction numerator 3 over denominator 1 half times scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end fraction end cell equals 5 row cell y plus fraction numerator 3 over denominator 1 half times y end fraction end cell equals 5 row cell y plus 6 over y end cell equals 5 row cell y squared plus 6 end cell equals cell 5 y end cell row cell y squared minus 5 y plus 6 end cell equals 0 row cell left parenthesis y minus 2 right parenthesis left parenthesis y minus 3 right parenthesis end cell equals 0 row cell y minus 2 end cell equals cell 0 blank atau blank y minus 3 equals 0 end cell row y equals cell 2 blank atau blank y equals 3 end cell end table$

Didapatkan hasil $y subscript 1 equals 2$ dan $y subscript 2 equals 3$ , maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 end cell equals 2 row cell scriptbase log space invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end cell equals 2 row cell 3 squared end cell equals cell x plus 1 end cell row 9 equals cell x plus 1 end cell row cell 9 minus 1 end cell equals x row 8 equals x end table$

$table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 2 end cell equals 3 row cell scriptbase log space invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 end cell equals 3 row cell 3 cubed end cell equals cell x plus 1 end cell row 27 equals cell x plus 1 end cell row cell 27 minus 1 end cell equals x row 26 equals x end table$

Didapatkan hasil $x subscript 1 equals 8$ dan $x subscript 2 equals 26$ , sehingga hasil dari $x subscript 1 plus x subscript 2$ adalah

$table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell 8 plus 26 end cell row blank equals 34 end table$

Jadi, dapat disimpulkan bahwa persamaan logaritma: $scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 3 plus fraction numerator 3 over denominator scriptbase log invisible function application left parenthesis x plus 1 right parenthesis end scriptbase presuperscript 9 end fraction equals 5$ mempunyai akar-akar $x subscript 1$ dan $x subscript 2$ maka $x subscript 1 plus x subscript 2 equals 34$ .

Oleh karena itu, jawaban yang benar adalah A.

Roboguru

Tentukan penyelesaian persamaan berikut. a.

Pembahasan Soal:

Diketahui $log presuperscript 5 space open parentheses x squared minus 9 x plus 14 close parentheses equals log presuperscript 5 space open parentheses negative 17 x plus 62 close parentheses$ maka:

$table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell table row blank cell log presuperscript 5 space open parentheses x squared minus 9 x plus 14 close parentheses equals log presuperscript 5 space open parentheses negative 17 x plus 62 close parentheses end cell row left right double arrow cell x squared minus 9 x plus 14 equals negative 17 x plus 62 end cell row left right double arrow cell x squared minus 9 x plus 17 x plus 14 minus 62 equals 0 end cell row left right double arrow cell x squared plus 8 x minus 48 equals 0 end cell row left right double arrow cell open parentheses x plus 12 close parentheses open parentheses x minus 4 close parentheses equals 0 end cell end table end cell end table$

diperoleh nilai yang memenuhi $x equals negative 12$ atau $x equals 4$

Syarat numerus 1)

$table row blank cell space space x squared minus 9 x plus 14 greater than 0 end cell row left right double arrow cell open parentheses x minus 2 close parentheses open parentheses x minus 7 close parentheses greater than 0 end cell row left right double arrow cell space space x less than 2 space atau space x greater than 7 end cell end table$

syarat numerus 2)

$table row blank cell negative 17 x plus 62 greater than 0 end cell row left right double arrow cell negative 17 x greater than negative 62 end cell row left right double arrow cell x less than 62 over 17 end cell end table$

Berdasarkan syarat numerus tersebut dengan demikian himpunan penyelesaian penyelesaian persamaan tersebut adalah $x equals negative 12$ .

Roboguru

Diketahui bahwa 3logx⋅6logx⋅9logx=3logx⋅6logx+3logx⋅9logx+6logx⋅9logx Tentukan nilai x (ada dua jawaban)

Pembahasan Soal:

Ingat

$^{a^{m}} lo g b = \frac{1}{m} \cdot^{a} lo g b$
$^{a} lo g b = \frac{^{c} lo g b}{^{c} lo g a}$
$^{a} lo g b = \frac{1}{^{b} lo g a}$
$^{a} lo g f (x) = b \Rightarrow f (c) = a^{c}$

Perhatikan perhitungan berikut

$^{3} lo g x \cdot^{6} lo g x \cdot^{9} lo g x =^{3} lo g x \cdot^{6} lo g x +^{3} lo g x \cdot^{9} lo g x +^{6} lo g x \cdot^{9} lo g x^{3} lo g x \cdot^{6} lo g x \cdot^{3^{2}} lo g x =^{3} lo g x \cdot^{6} lo g x +^{3} lo g x \cdot^{3^{2}} lo g x +^{6} lo g x \cdot^{3^{2}} lo g x \frac{1}{2} \cdot^{3} lo g x \cdot^{6} lo g x \cdot^{3} lo g x =^{3} lo g x \cdot^{6} lo g x + \frac{1}{2} \cdot^{3} lo g x \cdot^{3} lo g x + \frac{1}{2} \cdot^{6} lo g x \cdot^{3} lo g x \frac{1}{2} \cdot (^{3} lo g x)^{2} \cdot^{6} lo g x =^{3} lo g x \cdot^{6} lo g x + \frac{1}{2} \cdot (^{3} lo g x)^{2} + \frac{1}{2} \cdot^{6} lo g x \cdot^{3} lo g x \frac{1}{2} \cdot^{3} (lo g x)^{2} \cdot^{6} lo g x - \frac{1}{2} \cdot (^{3} lo g x)^{2} =^{3} lo g x \cdot^{6} lo g x + \frac{1}{2} \cdot^{6} lo g x \cdot^{3} lo g x \frac{1}{2} \cdot^{3} (lo g x)^{2} \cdot^{6} lo g x - \frac{1}{2} \cdot (^{3} lo g x)^{2} = \frac{3}{2} \cdot^{3} lo g x \cdot^{6} lo g x \frac{1}{2} \cdot^{3} (lo g x)^{2} \cdot \frac{^{3} l o g x}{^{3} l o g 6} - \frac{1}{2} \cdot (^{3} lo g x)^{2} = \frac{3}{2} \cdot^{3} lo g x \cdot \frac{^{3} l o g x}{^{3} l o g 6}^{3} (lo g x)^{2} \cdot \frac{^{3} l o g x}{^{3} l o g 6} - (^{3} lo g x)^{2} = 3 \cdot^{3} lo g x \cdot \frac{^{3} l o g x}{^{3} l o g 6} \frac{1}{^{3} l o g 6} \cdot (^{3} lo g x)^{3} - (^{3} lo g x)^{2} = \frac{3}{^{3} l o g 6} \cdot (^{3} lo g x)^{2} \frac{1}{^{3} l o g 6} \cdot (^{3} lo g x)^{3} - (^{3} lo g x)^{2} - \frac{3}{^{3} l o g 6} \cdot (^{3} lo g x)^{2} = 0 (^{3} lo g x)^{2} (\frac{^{3} l o g x}{^{3} l o g 6} - 1 - \frac{3}{^{3} l o g 6}) = 0 (^{3} lo g x)^{2} (^{6} lo g 3 \cdot^{3} lo g x - 1 - 3 \cdot^{6} lo g 3 \cdot) = 0$

Dari persamaan tersebut, diperoleh

$(^{3} lo g x)^{2}^{3} lo g x = 0 x x = = = 0 3^{0} 1$

atau

$^{6} lo g 3 \cdot^{3} lo g x - 1 - 3 \cdot^{6} lo g 3^{6} lo g 3 \cdot^{3} lo g x = 1 + 3 \cdot^{6} lo g 3^{3} lo g x = \frac{^{6} l o g 6 + ^{6} l o g 3 ^{3}}{^{6} l o g 3}^{3} lo g x = \frac{^{6} l o g 162}{^{6} l o g 3}^{3} lo g x =^{3} lo g 162 x = = 0162$