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Penyelesaian2log(x−3)+2log(x+3)≥4 adalah ...

Pertanyaan

Penyelesaian2log(x3)+2log(x+3)4 adalah ... 

  1. x5 

  2. x3 

  3. 3<x<3 

  4. 3<x5 

  5. 3x5 

Pembahasan Soal:

Ingat 

alogb+alogc=alogbc 

Perhatikan perhitungan berikut 

2log[(x3)(x+3)]2log162log[x29]2log16x29x225(x+5)(x5)2log(x3)+2log(x+3)2log161600  

Pembuat nol 

x+5xx5x====0505 

Uji x=0 

(x+5)(x5)===(0+5)(05)(5)(5)25negatif 

 

Syarat numerus 

x3xx+3x>>>>0303 


 


Jadi, HP={xx5,xR} 

Oleh karena itu, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 12 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Tentukan himpunan penyelesaian dari pertidaksamaan logaritma berikut ini: 2.

Pembahasan Soal:

Gunakan sifat:

begin mathsize 14px style log presuperscript a space b greater than log presuperscript a space c rightwards double arrow b greater than c log presuperscript a space b plus log presuperscript a space c equals log presuperscript a space b c end style

Syarat:

begin mathsize 14px style a greater than 0 comma space a not equal to 1 b comma space c greater than 0 end style 

Perhatikan perhitungan berikut:

begin mathsize 14px style table attributes columnalign right center left columnspacing 2px end attributes row cell log space x plus log space open parentheses x plus 9 close parentheses end cell greater than cell log space 10 end cell row cell log space open parentheses x open parentheses x plus 9 close parentheses close parentheses end cell greater than cell log space 10 end cell row cell log space open parentheses x squared plus 9 x close parentheses end cell greater than cell log space 10 end cell row cell x squared plus 9 x end cell greater than 10 row cell x squared plus 9 x minus 10 end cell greater than 0 row cell open parentheses x plus 10 close parentheses open parentheses x minus 1 close parentheses end cell greater than 0 end table x less than negative 10 space atau space x greater than 1 end style 

Berdasarkan syarat begin mathsize 14px style b comma space c greater than 0 end style, maka diperoleh perhitungan sebagai berikut:

begin mathsize 14px style table row cell left parenthesis x open parentheses x plus 9 close parentheses right parenthesis end cell greater than 0 end table x greater than 0 space atau space x plus 9 greater than 0 x greater than 0 space atau space x greater than negative 9 end style 

 

Dengan demikian, himpunan penyelesaiannya adalah begin mathsize 14px style left curly bracket x vertical line blank x greater than 1 comma x element of straight real numbers right curly bracket end style

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Roboguru

Pembahasan Soal:

log presuperscript 36 space left parenthesis x minus 4 right parenthesis plus log presuperscript 36 space left parenthesis x plus 1 right parenthesis less or equal than 1 half  S y a r a t space  x minus 4 greater than 0 rightwards double arrow x greater than 4 horizontal ellipsis horizontal ellipsis left parenthesis 1 right parenthesis  x plus 1 greater than 0 rightwards double arrow x greater than negative 1 horizontal ellipsis horizontal ellipsis left parenthesis 2 right parenthesis    K e m u d i a n space  log presuperscript 36 space left parenthesis x minus 4 right parenthesis plus log presuperscript 36 space left parenthesis x plus 1 right parenthesis less or equal than 1 half  log presuperscript 36 space left parenthesis x minus 4 right parenthesis plus log presuperscript 36 space left parenthesis x plus 1 right parenthesis less or equal than log presuperscript 36 space 36 to the power of 1 half end exponent  log presuperscript 36 space left parenthesis x minus 4 right parenthesis plus log presuperscript 36 space left parenthesis x plus 1 right parenthesis less or equal than log presuperscript 36 space 6  space left parenthesis x minus 4 right parenthesis plus left parenthesis x plus 1 right parenthesis less or equal than 6  x squared minus 3 x minus 10 less or equal than 0  x squared minus 3 x minus 10 less or equal than 0  minus 2 less or equal than x less or equal than 5 horizontal ellipsis horizontal ellipsis. left parenthesis 3 right parenthesis  I r i s k a n space s y a r a t space p e r t a m a space k e d u a space d a n space k e t i g a space colon

M a k a space d i d a p a t space 4 less than x less or equal than 5

0

Roboguru

Penyelesaian pertidaksamaan 2log x +  2log(x – 1) &lt; 1 adalah ….

Pembahasan Soal:

2log x +  2log(x – 1) < 1

Syarat :

x > 0 …..(1)

x – 1 > 0   x > 1 …..(2)

Kemudian

2log x +  2log(x – 1) < 1

2log x(x – 1) <  2log 2

x(x – 1) < 2

x(x – 1) < 0

x2 x – 2 <  0

(x – 2)(x + 1) < 0

–1 < x < 2 .....(3)

 

Iriskan syarat pertama kedua dan ketiga :

Maka didapat 1 < x < 2

 

0

Roboguru

Jika , nilai  yang memenuhi adalah ...

Pembahasan Soal:

Ingat sifat bentuk logaritma:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript a space b plus log presuperscript a space c end cell equals cell log presuperscript a space b c end cell row cell log presuperscript a space a end cell equals 1 row cell a to the power of scriptbase log space b end scriptbase presuperscript a end exponent end cell equals b row cell a to the power of n end cell equals cell b left right double arrow log presuperscript a space b equals n end cell end table  

Ingat sifat bilangan berpangkat:

table attributes columnalign right center left columnspacing 0px end attributes row cell x to the power of m plus n end exponent end cell equals cell x to the power of m times x to the power of n end cell row cell open parentheses x to the power of m close parentheses to the power of n end cell equals cell x to the power of m n end exponent end cell row cell x to the power of negative m end exponent end cell equals cell 1 over x to the power of m end cell row cell open parentheses x y close parentheses to the power of m end cell equals cell x to the power of m y to the power of m end cell end table  

dan ingat pada pertidaksamaan logaritma untuk a greater than 1berlaku:

log presuperscript a space f open parentheses x close parentheses greater than log presuperscript a space g open parentheses x close parentheses rightwards arrow f open parentheses x close parentheses greater than g open parentheses x close parentheses comma space f open parentheses x close parentheses greater than 0 comma space g open parentheses x close parentheses greater than 0  

Sehingga

table row blank cell open parentheses 2 x close parentheses to the power of 1 plus log presuperscript 2 space 2 x end exponent greater than 64 x cubed end cell row blank cell open parentheses 2 x close parentheses to the power of 1 plus log presuperscript 2 space 2 plus log presuperscript 2 space x end exponent greater than 2 to the power of 6 x cubed end cell row blank cell open parentheses 2 x close parentheses to the power of 1 plus 1 plus log presuperscript 2 space x end exponent greater than 2 squared times 2 to the power of 4 times x times x squared end cell row blank cell open parentheses 2 x close parentheses to the power of 2 plus log presuperscript 2 space x end exponent greater than open parentheses 2 x close parentheses squared times 2 to the power of 4 x end cell row blank cell open parentheses 2 x close parentheses squared times open parentheses 2 x close parentheses to the power of log presuperscript 2 space x end exponent greater than open parentheses 2 x close parentheses squared times 2 to the power of 4 x end cell row blank cell open parentheses 2 x close parentheses to the power of log presuperscript 2 space x end exponent greater than 2 to the power of 4 x end cell row blank cell 2 to the power of log presuperscript 2 space x end exponent times x to the power of log presuperscript 2 space x end exponent greater than 2 to the power of 4 x end cell row blank cell x times x to the power of log presuperscript 2 space x end exponent greater than 2 to the power of 4 x end cell row blank cell x to the power of log presuperscript 2 space x end exponent greater than 2 to the power of 4 end cell end table 

Misalkan log presuperscript 2 space x equals p maka 2 to the power of p equals x sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell x to the power of log presuperscript 2 space x end exponent end cell greater than cell 2 to the power of 4 end cell row cell open parentheses 2 to the power of p close parentheses to the power of p end cell greater than cell 2 to the power of 4 end cell row cell 2 to the power of p squared end exponent end cell greater than cell 2 to the power of 4 end cell row cell p squared end cell greater than 4 row cell p squared minus 4 end cell greater than 0 row cell open parentheses p minus 2 close parentheses open parentheses p plus 2 close parentheses end cell greater than 0 end table 

sehingga nilai p yang memenuhi adalah p greater than 2 atau p less than negative 2.

Untuk p greater than 2 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space x end cell greater than 2 row cell log presuperscript 2 space x end cell greater than cell log presuperscript 2 space 2 squared end cell row x greater than cell 2 squared end cell row x greater than 4 end table 

 dan untuk nilai p less than negative 2 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space x end cell less than cell negative 2 end cell row cell log presuperscript 2 space x end cell less than cell log presuperscript 2 space 2 to the power of negative 2 end exponent end cell row x less than cell 2 to the power of negative 2 end exponent end cell row x less than cell 1 fourth end cell end table  

Syarat numerus:

 2 x greater than 0 rightwards arrow x greater than 0   

Diperoleh x greater than 0x less than 1 fourth, dan x greater than 4 maka irisannya adalah 0 less than x less than 1 fourth atau x greater than 4. Dengan demikian nilai x yang memenuhi adalah 0 less than x less than 1 fourth atau x greater than 4.

Oleh karena itu, jawaban yang benar adalah E.

 

0

Roboguru

Nilai yang memenuhi pertidaksamaan adalah ...

Pembahasan Soal:

Ingatlah syarat pertidaksamaan logaritma dengan bilangan pokok (basis) a greater than 1, yaitu:

log presuperscript a space f open parentheses x close parentheses less or equal than log presuperscript a space g open parentheses x close parentheses space rightwards arrow space f left parenthesis x right parenthesis less or equal than g open parentheses x close parentheses

dimana f left parenthesis x right parenthesis greater than 0 dan g left parenthesis x right parenthesis greater than 0.

Berdasarkan hal tersebut, maka pertidaksamaan harus memenuhi 3 syarat:

  • log presuperscript a space f open parentheses x close parentheses less or equal than log presuperscript a space g open parentheses x close parentheses space rightwards arrow space f left parenthesis x right parenthesis less or equal than g open parentheses x close parentheses

table attributes columnalign right center left columnspacing 0px end attributes row cell log open parentheses x minus 3 close parentheses plus log open parentheses x plus 8 close parentheses end cell less than cell log open parentheses 2 x plus 4 close parentheses end cell row cell log space open parentheses x minus 3 close parentheses open parentheses x plus 8 close parentheses end cell less than cell log open parentheses 2 x plus 4 close parentheses end cell row cell log open parentheses x squared plus 5 x minus 24 close parentheses end cell less than cell log open parentheses 2 x plus 4 close parentheses end cell row cell x squared plus 5 x minus 24 end cell less than cell 2 x plus 4 end cell row cell x squared plus 5 x minus 2 x minus 24 minus 4 end cell less than 0 row cell x squared plus 3 x minus 28 end cell less than 0 row cell open parentheses x plus 7 close parentheses open parentheses x minus 4 close parentheses end cell less than 0 end table

lakukan uji titik pada garis bilangan sehingga didapat solusi:

H P subscript 1 equals open curly brackets negative 7 less than x less than 4 close curly brackets

  • f left parenthesis x right parenthesis greater than 0

table attributes columnalign right center left columnspacing 0px end attributes row cell x squared plus 5 x minus 24 end cell greater than 0 row cell open parentheses x plus 8 close parentheses open parentheses x minus 3 close parentheses end cell greater than 0 end table

lakukan uji titik pada garis bilangan sehingga didapat solusi:

H P subscript 2 equals open curly brackets x less than negative 8 space atau space x greater than 3 close curly brackets

  • g left parenthesis x right parenthesis greater than 0

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x plus 4 end cell greater than 0 row cell 2 x end cell greater than cell negative 4 end cell row x greater than cell negative 2 end cell end table

H P subscript 3 equals open curly brackets x greater than negative 2 close curly brackets

Selanjutnya, iriskan solusi ketiga syarat H P subscript 1 equals open curly brackets negative 7 less than x less than 4 close curly brackets, H P subscript 2 equals open curly brackets x less than negative 8 space atau space x greater than 3 close curly brackets, dan H P subscript 3 equals open curly brackets x greater than negative 2 close curly brackets sehingga:

Dengan demikian, nilai x yang memenuhi pertidaksamaan logaritma tersebut adalah 3 less than x less than 4.

Oleh karena itu, jawaban yang benar adalah D.

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