Persamaan logaritma  mempunyai penyelesaian  dan x2​. Tentukan nilai x1​+x2​.

Pertanyaan

Persamaan logaritma open parentheses log presuperscript 4 space x close parentheses squared minus log presuperscript 2 space square root of x minus 3 over 4 equals 0 mempunyai penyelesaian x subscript 1 dan x subscript 2. Tentukan nilai x subscript 1 plus x subscript 2.

S. Amamah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

Ingat sifat-sifat logaritma:

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

Diketahui open parentheses log presuperscript 4 space x close parentheses squared minus log presuperscript 2 space square root of x minus 3 over 4 equals 0 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses log presuperscript 4 space x close parentheses squared minus log presuperscript 2 space square root of x minus 3 over 4 end cell equals 0 row cell open parentheses log presuperscript 2 squared end presuperscript space x close parentheses squared minus log presuperscript 2 space x to the power of begin inline style 1 half end style end exponent minus 3 over 4 end cell equals 0 row cell open parentheses 1 half times log presuperscript 2 space x close parentheses squared minus 1 half times log presuperscript 2 space x minus 3 over 4 end cell equals 0 row cell 1 fourth open parentheses log presuperscript 2 space x close parentheses squared minus 1 half times log presuperscript 2 space x minus 3 over 4 end cell equals 0 row blank blank blank end table 

misalkan log presuperscript 2 space x equals a maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 1 fourth a squared minus 1 half a minus 3 over 4 end cell equals 0 row cell a squared minus 2 a minus 3 end cell equals 0 row cell open parentheses a minus 3 close parentheses open parentheses a plus 1 close parentheses end cell equals 0 end table 

diperoleh a equals 3 atau a equals negative 1.

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

Untuk a equals 3 maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space x end cell equals 3 row cell log presuperscript 2 space x end cell equals cell log presuperscript 2 space 2 cubed end cell row cell log presuperscript 2 space x end cell equals cell space log presuperscript 2 space 8 end cell row x equals 8 end table 

untuk maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space x end cell equals cell negative 1 end cell row cell log presuperscript 2 space x end cell equals cell log presuperscript 2 space 2 to the power of negative 1 end exponent end cell row cell log presuperscript 2 space x end cell equals cell space log presuperscript 2 space begin inline style 1 half end style end cell row x equals cell begin inline style 1 half end style end cell end table 

Maka nilai x subscript 1 plus x subscript 2:

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 1 half end cell row blank equals cell 8 1 half end cell row blank equals cell 17 over 2 end cell end table 

Dengan demikian nilai x subscript 1 plus x subscript 2 adalah 17 over 2.

 

48

0.0 (0 rating)

Pertanyaan serupa

Jika 16log (x−2)−16log (x2−4x+4)1​=−2, tentukan nilai x yang memenuhi persamaan tersebut.

34

0.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Produk Lainnya

Fitur Roboguru

Topik Roboguru

Hubungi Kami

Ruangguru WhatsApp

081578200000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2022 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia