Roboguru

Pertanyaan

Himpunan penyelesaian pertidaksamaan log presuperscript 2 space left parenthesis x plus 2 right parenthesis plus log presuperscript 2 left parenthesis x minus 2 right parenthesis less or equal than log presuperscript 2 5 adalah...

  1. open curly brackets right enclose x x greater or equal than negative 2 close curly brackets

  2. open curly brackets right enclose x x greater or equal than 2 close curly brackets

  3. open curly brackets right enclose x x greater or equal than 3 close curly brackets

  4. open curly brackets right enclose x 2 less than x less or equal than 3 close curly brackets

  5. open curly brackets right enclose x minus 2 less than x less than 2 close curly brackets

F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Pembahasan

log presuperscript 2 space left parenthesis x plus 2 right parenthesis plus log presuperscript 2 space left parenthesis x minus 2 right parenthesis less or equal than log presuperscript 2 space 5  S y a r a t space left parenthesis x plus 2 right parenthesis greater than 0 rightwards arrow space x greater than negative 2  space space space space space space space space space space space space space left parenthesis x minus 2 right parenthesis greater than 0 rightwards arrow space x greater than 2  log presuperscript 2 space left parenthesis x plus 2 right parenthesis plus log presuperscript 2 space left parenthesis x minus 2 right parenthesis less or equal than log presuperscript 2 space 5  log presuperscript 2 space left parenthesis x plus 2 right parenthesis left parenthesis x minus 2 right parenthesis less or equal than log presuperscript 2 space 5  log presuperscript 2 space left parenthesis x squared minus 4 right parenthesis less or equal than log presuperscript 2 space 5  x squared minus 4 less or equal than 5  x squared minus 9 less or equal than 0  left parenthesis x plus 3 right parenthesis left parenthesis x minus 3 right parenthesis less or equal than 0  x equals negative 3 space a t a u space x equals 3 space  U j i space t i t i k space u n t u k space x equals 1 comma m a k a space 1 squared minus 9 equals negative 8 less than 0  K a r e n a space minus 3 less than 1 less than 3 comma space m a k a space u n t u k space d a e r a h space minus 3 less than x less than 3 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

dengan syarat : x > -2 dan x > 2, maka 

Sehingga, irisan area dari gambar di atas adalah

HP = open curly brackets right enclose x 2 less than x less or equal than 3 close curly brackets

35

0.0 (0 rating)

Pertanyaan serupa

Himpunan penyelesaian dari pertidaksamaan 5log(x−3)+5log(x+1)≤1 adalah...

434

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 Ruangguru

Produk Ruangguru

Produk Lainnya

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