Roboguru

Tentukan nilai dari 2log81​+3log91​+5log1251​

Pertanyaan

Tentukan nilai dari 2log81+3log91+5log1251  

Pembahasan Soal:

Untuk pertidaksamaan bentuk f(x)g(x) maka ruas kiri sudah pasti bernilai positif namun ruas kanan belum tentu bernilai positif, sehingga bentuk pertidaksamaan seperti ini perlu diuraikan menjadi dua kemungkinan yaitu g(x)0dang(x)<0

Kemungkinan Kasus 1 : x3<0x>3

Olehkarena3x+10danx<3maka3x+1x3terpenuhi untuk semua x.

Syarat akar : 

 3x+1x031

Irisan dari x<3,xR,danx31 direpresentasikan oleh garais bilangan berikut :

Sehingga HP kasus 1 :31x<3 

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Septianingsih

Mahasiswa/Alumni Universitas Gadjah Mada

Terakhir diupdate 30 Agustus 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

Tentukan nilai dari 2log81​+3log91​+5log1251​

Pembahasan Soal:

Gunakan sifat logaritma 

alogaalogan==1naloga 

Maka pada soal berlaku :

 =====2log81+3log91+5log12512log231+3log321+5log5312log23+3log32+5log53(32log2)+(23log3)+(35log5)(3)+(2)+(3)8 

Dengan demikian nilai dari 2log81+3log91+5log1251=8

0

Roboguru

Jika , nilai

Pembahasan Soal:

Ingat sifat-sifat logaritma 

table attributes columnalign right center left columnspacing 0px end attributes row cell blank to the power of a log space b c equals to the power of a log space b plus to the power of a log space c to the power of a log space b over c equals to the power of a log space b minus to the power of a log space c to the power of a log space b to the power of m end cell equals cell m cross times to the power of a log space b end cell end table 

Maka dengan menggunakan sifat-sifat tersebut, diperoleh 

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses 2 a close parentheses plus f open parentheses 2 over a close parentheses end cell equals cell fraction numerator space to the power of 4 log space open parentheses 2 a close parentheses over denominator 1 minus 2 cross times to the power of 4 log open parentheses 2 a close parentheses end fraction plus fraction numerator space to the power of 4 log space open parentheses begin display style 2 over a end style close parentheses over denominator 1 minus 2 cross times to the power of 4 log open parentheses begin display style 2 over a end style close parentheses end fraction end cell row blank equals cell fraction numerator space to the power of 4 log space open parentheses 2 a close parentheses over denominator blank to the power of 4 log space 4 minus to the power of 4 log space open parentheses 2 a close parentheses squared end fraction plus fraction numerator space to the power of 4 log space open parentheses 2 over a close parentheses over denominator blank to the power of 4 log space 4 minus to the power of 4 log open parentheses 2 over a close parentheses squared end fraction end cell row blank equals cell fraction numerator space to the power of 4 log space open parentheses 2 a close parentheses over denominator blank to the power of 4 log space 4 minus to the power of 4 log space 4 a squared end fraction plus fraction numerator space to the power of 4 log space open parentheses 2 over a close parentheses over denominator blank to the power of 4 log space 4 minus to the power of 4 log open parentheses 4 over a squared close parentheses end fraction end cell row blank equals cell fraction numerator space to the power of 4 log space 2 a over denominator blank to the power of 4 log open parentheses begin display style fraction numerator 4 over denominator 4 a squared end fraction end style close parentheses end fraction plus fraction numerator space to the power of 4 log space open parentheses 2 over a close parentheses over denominator blank to the power of 4 log space open parentheses begin display style fraction numerator 4 over denominator 4 over a squared end fraction end style close parentheses end fraction end cell row blank equals cell fraction numerator space to the power of 4 log space 2 a over denominator blank to the power of 4 log open parentheses begin display style 1 over a squared end style close parentheses end fraction plus fraction numerator space to the power of 4 log space open parentheses 2 over a close parentheses over denominator blank to the power of 4 log space open parentheses begin display style a squared end style close parentheses end fraction end cell row blank equals cell fraction numerator space to the power of 4 log space 2 a over denominator negative to the power of 4 log space a squared end fraction plus fraction numerator space to the power of 4 log space open parentheses 2 over a close parentheses over denominator blank to the power of 4 log space begin display style a squared end style end fraction end cell row blank equals cell fraction numerator space minus to the power of 4 log space 2 a over denominator blank to the power of 4 log space a squared end fraction plus fraction numerator space to the power of 4 log space open parentheses 2 over a close parentheses over denominator blank to the power of 4 log space begin display style a squared end style end fraction end cell row blank equals cell fraction numerator space minus to the power of 4 log space 2 a plus to the power of 4 log space open parentheses 2 over a close parentheses over denominator blank to the power of 4 log space a squared end fraction end cell row blank equals cell fraction numerator space minus to the power of 4 log space 2 minus to the power of 4 log space a plus to the power of 4 log space 2 minus to the power of 4 log space a over denominator 2 cross times to the power of 4 log space a end fraction end cell row blank equals cell fraction numerator negative 2 to the power of 4 log space a over denominator 2 to the power of 4 log space a end fraction end cell row blank equals cell negative 1 end cell end table  

Jadi, f open parentheses 2 a close parentheses plus f open parentheses 2 over a close parentheses equals negative 1

Dengan demikian, jawaban yang tepat adalah B.

0

Roboguru

Nilai dari 3log5−3log15+3log9adalah.....

Pembahasan Soal:

Gunakan sifat logaritma :

alogcb=alogbalogc  

aloga=1 

alog(bc)=alogb+alogc

3log53log15+3log9===3log(155×9)3log31 

Dengan demikian nilai dari 3log53log15+3log9=1 

0

Roboguru

Nilai dari

Pembahasan Soal:

Sifat-sifat yang digunakan:

log presuperscript straight a space straight b space space plus space log presuperscript straight a space straight c space equals space log presuperscript straight a space left parenthesis straight b. straight c right parenthesis  log presuperscript straight a space straight b space minus space log presuperscript straight a space straight c space equals space log presuperscript straight a space straight b over straight c  log presuperscript straight a space straight b to the power of straight n equals straight n space log presuperscript straight a space straight b  log presuperscript straight a space straight a space equals space 1    log presuperscript 2 space 6 space plus space log presuperscript 2 space 8 space minus space log presuperscript 2 space 12 space equals space log presuperscript 2 space open parentheses fraction numerator 6.8 over denominator 12 end fraction close parentheses  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space log presuperscript 2 space 4  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space log presuperscript 2 space 2 squared  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2. space log presuperscript 2 space 2  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2.1 space  space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space 2

0

Roboguru

Resolusi  ketimpangan  adalah ....

Pembahasan Soal:

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