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Jika , nilai

Pertanyaan

Jika f open parentheses x close parentheses equals fraction numerator space to the power of 4 log space x over denominator 1 minus 2 cross times to the power of 4 log space x end fraction, nilai f open parentheses 2 a close parentheses plus f open parentheses 2 over a close parentheses equals... 

  1. negative a 

  2. negative 1 

  3. 0 

  4. 1 

  5. a 

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.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

L. Rante

Mahasiswa/Alumni Universitas Negeri Makassar

Terakhir diupdate 11 Juli 2021

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Pertanyaan yang serupa

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 

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:

Beberapa sifat logaritma:

  • blank to the power of a log space a equals 1
  • blank to the power of a log space b plus to the power of a log space c equals to the power of a log space b c
  • blank to the power of a log space b to the power of c equals c. to the power of a log space b

Sehingga soal di atas menjadi:

undefined

0

Roboguru

.

Pembahasan Soal:

Gunakan sifat bentuk logaritma berikut:

table attributes columnalign right center left columnspacing 2px end attributes row cell log presuperscript a space open parentheses b over c close parentheses end cell equals cell log presuperscript a space b minus log presuperscript a space c end cell row cell log presuperscript a space open parentheses b c close parentheses end cell equals cell log presuperscript a space b plus log presuperscript a space c end cell row cell log presuperscript a space open parentheses b to the power of m close parentheses end cell equals cell m space open parentheses log presuperscript a space b close parentheses end cell row cell log presuperscript a space a end cell equals 1 end table

Menggunakan sifat bentuk logaritma di atas, nilai log presuperscript 5 space 75 minus log presuperscript 5 space 24 plus log presuperscript 5 space 8 dapat diperoleh sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell log presuperscript 5 space 75 minus log presuperscript 5 space 24 plus log presuperscript 5 space 8 end cell equals cell log presuperscript 5 space open parentheses 75 over 24 close parentheses plus log presuperscript 5 space 8 end cell row blank equals cell log presuperscript 5 space open parentheses 75 over 24 cross times 8 close parentheses end cell row blank equals cell log presuperscript 5 space 25 end cell row blank equals cell log presuperscript 5 space 5 squared end cell row blank equals cell 2 space open parentheses log presuperscript 5 space 5 close parentheses end cell row blank equals cell 2 times 1 end cell row blank equals 2 end table

Diperoleh bahwa log presuperscript 5 space 75 minus log presuperscript 5 space 24 plus log presuperscript 5 space 8 equals 2.

Jadi, jawaban yang tepat adalah B.

0

Roboguru

Bentuk sederhana dari alog2x+3(alogx−alogy)alogx​+alogy​−21​alogyx​​ adalah...

Pembahasan Soal:

Ingat kembali sifat-sifat logaritma berikut ini:

alogbnalogb+alogcalogbalogc===nalogbalogbcalogcb

Sehingga diperoleh perhitungan:

alog2x+3(alogxalogy)alogx+alogy21alogyx==========alog2x+3(alogyx)alogx21+alogy2121alogyxalog2x+3(alogyx)21(alogx+alogyalogyx)alog2x+alog(yx)321(alogxyalogyx)alog2x(yx)321(alogyxxy)alogy32x421alogy2alogy32x4alogylogy32x4logyylogy32x4ylogyylog2x4ylogy31ylog2x431   

Jadi, bentuk sederhana dari logaritma di atas adalah ylog2x431

0

Roboguru

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