Roboguru

Himpunan penyelesaian pertidaksamaan2log(x+2)+2log(x−2)≤2log5 adalah ...

Pertanyaan

Himpunan penyelesaian pertidaksamaan2log(x+2)+2log(x2)2log5 adalah ... 

  1. {xx2}  

  2. {xx2} 

  3. {xx3} 

  4. {x∣2<x3} 

  5. {x2<x2} 

Pembahasan Soal:

Ingat 

alogbc=alogb+alogc 

Perhatikan perhitungan berikut 

2log[(x+2)(x2)]2log52log(x24)2log5x24x29(x+3)(x3)2log(x+2)+2log(x2)2log5500 

Pembuat nol 

x+3xx3x====0303 

Uji x=0 

(x+3)(x3)===(0+3)(03)(3)(3)9negatif 

Uji syarat numerus  

x+2xx2x>>>>0202 


 


Jadi, HP={x∣2<x3,xR}

Oleh karena itu, jawaban yang benar adalah D.

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

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

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.

0

Roboguru

Penyelesaian dari pertidaksamaan adalah ....

Pembahasan Soal:

table row 25 row space end table log open parentheses x minus 3 close parentheses plus table row 25 row space end table log open parentheses x plus 1 close parentheses less or equal than 1 half  S y a r a t space open parentheses x minus 3 close parentheses greater than 0 rightwards arrow x greater than 3  open parentheses x plus 1 close parentheses greater than 0 rightwards arrow x greater than negative 1  table row 25 row space end table log open parentheses x minus 3 close parentheses plus table row 25 row space end table log open parentheses x plus 1 close parentheses less or equal than 1 half  table row 25 row space end table log open parentheses x minus 3 close parentheses open parentheses x plus 1 close parentheses less or equal than 1 half table row 25 row space end table log 25  table row 25 row space end table log left parenthesis x squared minus 2 x minus 3 right parenthesis less or equal than table row 25 row space end table log 25 to the power of 1 half end exponent  x squared minus 2 x minus 3 less or equal than 5  x squared minus 2 x minus 8 less or equal than 0  left parenthesis x minus 4 right parenthesis left parenthesis x plus 2 right parenthesis less or equal than 0  x equals 4 space a t a u space x equals negative 2  U j i space t i t i k space u n t u k space x space equals space 0 comma space m a k a space 0 squared minus 2 open parentheses 0 close parentheses minus 3 equals negative 3 less or equal than 5 left parenthesis m e m e n u h i right parenthesis  K a r e n a space minus 2 less or equal than 0 less or equal than 4 comma space m a k a space u n t u k space d a e r a h space minus 2 less or equal than x less or equal than 4 space b e r n i l a i space n e g a t i f  A m b i l space y a n g space n e g a t i f comma space m a k a space d i d a p a t space p e n y e l e s a i a n

0

Roboguru

Tentukan himpunan penyelesaian dari pertidaksamaan logaritma berikut:

Pembahasan Soal:

Ingat bahwa

begin mathsize 14px style scriptbase log space f open parentheses x close parentheses plus scriptbase log space g open parentheses x close parentheses equals scriptbase log space f open parentheses x close parentheses times g open parentheses x close parentheses end scriptbase presuperscript a end scriptbase presuperscript a end scriptbase presuperscript a end style 

dan 

begin mathsize 14px style scriptbase log space b to the power of n equals n times scriptbase log space b end scriptbase presuperscript a end scriptbase presuperscript a end style.

Sehingga diperoleh pertidaksamaan logaritma:

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell table attributes columnalign right center left end attributes row cell scriptbase log open parentheses x minus 3 close parentheses plus scriptbase log open parentheses x plus 3 close parentheses end scriptbase presuperscript 2 end scriptbase presuperscript 2 end cell greater or equal than 4 row cell scriptbase log open parentheses x minus 3 close parentheses open parentheses x plus 3 close parentheses end scriptbase presuperscript 2 end cell greater or equal than cell 4 times scriptbase log space 2 end scriptbase presuperscript 2 end cell row cell scriptbase log open parentheses x squared minus 9 close parentheses end scriptbase presuperscript 2 end cell greater or equal than cell scriptbase log space 2 to the power of 4 end scriptbase presuperscript 2 end cell row cell x squared minus 9 end cell greater or equal than 16 row cell x squared minus 9 minus 16 end cell greater or equal than 0 row cell x squared minus 25 end cell greater or equal than 0 row cell open parentheses x minus 5 close parentheses open parentheses x plus 5 close parentheses end cell greater or equal than 0 row cell x greater or equal than 5 end cell logical or cell x greater or equal than negative 5 end cell end table end cell row blank blank blank end table

Garis bilangannya adalah sebagai berikut:

Syarat numerus

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x close parentheses end cell greater than 0 row cell x minus 3 end cell greater than 0 row x greater than 3 end table end style   dan   begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g open parentheses x close parentheses end cell greater than 0 row cell x plus 3 end cell greater than 0 row x greater than cell negative 3 end cell end table end style

Garis bilangannya menjadi:

Dengan demikian, diperoleh begin mathsize 14px style HP equals open curly brackets right enclose x space x greater or equal than 5 space close curly brackets end style.

0

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

Tentukan himpunan  penyelesaian dari log∣x+1∣≥log3+log∣2x−1∣!

Pembahasan Soal:

Ingat 

  • alogf(x)alogg(x)f(x)g(x);a>1 
  • f(x)={f(x),f(x)0f(x),f(x)<0 

Perhatikan perhitungan berikut 

x+12x1=={(x+1),x1(x+1),x<1{(2x1),x21(2x1),x<21  

Untuk x<1danx21 tidak ada x yang memenuhi 

Untuk 1<x<21  

logx+1log(x+1)log(x+1)log[(2x1)]log12xx+112xx+112xx+1312xx+112x36x12x7x2log3+log2x1log3+log[(2x1)]log3log33000  

Pembuat nol 

7x2x12xx====072021 

Uji x=0 

12x7x2==12(0)7(0)27positif 


 

Jadi HP={x72<x<21,xR}.  

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved