Roboguru

Nilai x yang memenuhi pertidaksamaan:   adalah....

Pertanyaan

Nilai x yang memenuhi pertidaksamaan:

scriptbase l og space left parenthesis x plus square root of 24 space right parenthesis end scriptbase presuperscript 1 fourth end presuperscript space greater than negative scriptbase log space left parenthesis x minus square root of 24 right parenthesis end scriptbase presuperscript 1 fourth end presuperscript  adalah....

Pembahasan Soal:

Pretama kita sederhanakan terlebih dahulu bentuk pertidaksamaan logaritma tersebut:

scriptbase l og space left parenthesis x plus square root of 24 space right parenthesis end scriptbase presuperscript 1 fourth end presuperscript space greater than negative scriptbase log space left parenthesis x minus square root of 24 right parenthesis end scriptbase presuperscript 1 fourth end presuperscript scriptbase l og space left parenthesis x plus square root of 24 space right parenthesis end scriptbase presuperscript 1 fourth end presuperscript space greater than scriptbase log space left parenthesis x minus square root of 24 right parenthesis to the power of negative 1 end exponent end scriptbase presuperscript 1 fourth end presuperscript 

Syarat pertidaksamaan logaritma, bentuk

log presuperscript straight a space straight f left parenthesis straight x right parenthesis greater than log presuperscript straight a space straight g open parentheses straight x close parentheses Jika colon space 0 less than straight a less than 1 space maka space straight f left parenthesis straight x right parenthesis less than space straight g open parentheses straight x close parentheses space dengan space syarat space straight f left parenthesis straight x right parenthesis greater than 0 space dan space space straight g open parentheses straight x close parentheses greater than 0   

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus square root of 24 end cell less than cell fraction numerator 1 over denominator x minus square root of 24 end fraction end cell row cell x squared minus 24 end cell less than 0 row x less than cell plus-or-minus square root of 24 end cell row blank blank cell x less than negative square root of 24 space atau space x less than square root of 24 end cell end table

dengan syarat:

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus square root of 24 end cell greater than 0 row x greater than cell negative square root of 24 end cell end table 

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 1 over denominator x minus square root of 24 end fraction end cell greater than 0 row x not equal to cell square root of 24 end cell end table

Sehingga pada garis bilangan:

Jadi, nilai x yang memenuhi pertidaksamaan adalah x less than square root of 24

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 02 Juni 2021

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