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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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Pertanyaan yang serupa

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 1f 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.

 

0

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.

0

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.

0

Roboguru

Tentukan penyelesaian persamaan berikut. a.

Pembahasan Soal:

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 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.

 

0

Roboguru

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

Pembahasan Soal:

Ingat 

  • amlogb=m1alogb
  •  alogb=clogaclogb 
  • alogb=bloga1 
  • alogf(x)=bf(c)=ac 

Perhatikan perhitungan berikut 

3logx6logx9logx=3logx6logx+3logx9logx+6logx9logx3logx6logx32logx=3logx6logx+3logx32logx+6logx32logx213logx6logx3logx=3logx6logx+213logx3logx+216logx3logx21(3logx)26logx=3logx6logx+21(3logx)2+216logx3logx213(logx)26logx21(3logx)2=3logx6logx+216logx3logx213(logx)26logx21(3logx)2=233logx6logx213(logx)23log63logx21(3logx)2=233logx3log63logx3(logx)23log63logx(3logx)2=33logx3log63logx3log61(3logx)3(3logx)2=3log63(3logx)23log61(3logx)3(3logx)23log63(3logx)2=0(3logx)2(3log63logx13log63)=0(3logx)2(6log33logx136log3)=0 

Dari persamaan tersebut, diperoleh 

(3logx)23logx=0xx===0301 

atau 

6log33logx136log36log33logx=1+36log33logx=6log36log6+6log333logx=6log36log1623logx=3log162x==0162  

Dengan demikian, nilai x yang memenuhi persamaan di tersebut adalah  x=0ataux=162

0

Roboguru

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