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Tentukan nilai logaritma berikut. a. 3log27+3log6−3log2

Pertanyaan

Tentukan nilai logaritma berikut.

a. 3log27+3log63log2

Pembahasan Soal:

Ingat sifat-sifat logaritma berikut:

  • alogb+alogc=alogbc
  • alogbalogc=alogcb
  • alogan===nalogan1n

Berdasarkan sifat tersebut, maka

3log27+3log63log2=====3log22763log2733log3333log344

Jadi, diperoleh nilai 3log27+3log63log2=4.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

H. Janatu

Mahasiswa/Alumni Universitas Riau

Terakhir diupdate 12 September 2021

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Pertanyaan yang serupa

Hasil  adalah. . . .

Pembahasan Soal:

Dengan menggunakan sifat bentuk logaritma berikut.

Untuk a comma b comma c greater than 0 dan a not equal to 1, berlaku:

  1. log presuperscript a space a equals 1
  2. log presuperscript a space b to the power of m equals m open parentheses log presuperscript a space b close parentheses
  3. log presuperscript a b over c equals log presuperscript a space b minus log presuperscript a space c
  4. log presuperscript a open parentheses b cross times c close parentheses equals log presuperscript a space b cross times log presuperscript a space c

hasil dari log presuperscript 2 space 18 minus log presuperscript 2 space 8 square root of 2 plus log presuperscript 2 space 1 over 27 minus log presuperscript 2 space 1 third sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell log presuperscript 2 space 18 minus log presuperscript 2 space 8 square root of 2 plus log presuperscript 2 space 1 over 27 minus log presuperscript 2 space 1 third end cell row blank equals cell log presuperscript 2 open parentheses fraction numerator 18 over denominator 8 square root of 2 end fraction close parentheses plus log presuperscript 2 space 1 over 27 minus log presuperscript 2 space 1 third space space space space space space space space space space space space space space Sifat space 3 end cell row blank equals cell log presuperscript 2 open parentheses fraction numerator 18 over denominator 8 square root of 2 end fraction cross times 1 over 27 close parentheses minus log presuperscript 2 space 1 third space space space space space space space space space space space space space space space space space space space space space space Sifat space 4 end cell row blank equals cell log presuperscript 2 open parentheses fraction numerator 18 over denominator 8 square root of 2 end fraction cross times 1 over 27 divided by 1 third 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 Sifat space 3 end cell row blank equals cell log presuperscript 2 open parentheses fraction numerator 18 over denominator 8 square root of 2 end fraction cross times 1 over 27 cross times 3 over 1 close parentheses end cell row blank equals cell log presuperscript 2 open parentheses up diagonal strike 18 to the power of up diagonal strike 2 to the power of 1 end exponent over up diagonal strike 8 square root of 2 end strike to the power of 4 square root of 2 end exponent cross times 1 over up diagonal strike 27 subscript up diagonal strike 9 subscript 1 end subscript cross times up diagonal strike 3 to the power of 1 close parentheses end cell row blank equals cell log presuperscript 2 fraction numerator 1 over denominator 4 square root of 2 end fraction end cell row blank equals cell log presuperscript 2 fraction numerator 1 over denominator 2 squared cross times 2 to the power of begin display style 1 half end style end exponent end fraction end cell row blank equals cell log presuperscript 2 1 over 2 to the power of open parentheses 2 plus begin display style 1 half end style close parentheses end exponent end cell row blank equals cell log presuperscript 2 1 over 2 to the power of begin display style 5 over 2 end style end exponent end cell row blank equals cell log presuperscript 2 space 2 to the power of open parentheses negative 5 over 2 close parentheses end exponent end cell row blank equals cell open parentheses negative 5 over 2 close parentheses cross times 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 Sifat space 2 end cell row blank equals cell open parentheses negative 5 over 2 close parentheses cross times 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 space space space space Sifat space 1 end cell row blank equals cell negative 5 over 2 end cell end table


Hasil dari log presuperscript 2 space 18 minus log presuperscript 2 space 8 square root of 2 plus log presuperscript 2 space 1 over 27 minus log presuperscript 2 space 1 third adalah negative 5 over 2.

Oleh karena itu, jawaban yang benar adalah A.

0

Roboguru

Nilai dari ....

Pembahasan Soal:

Gunakan sifat-sifat logaritma, yaitu :

log presuperscript a space straight a to the power of straight n equals straight n  log presuperscript straight a space straight b plus log presuperscript straight a space straight c equals log presuperscript straight a space bc  log presuperscript straight a space straight b minus log presuperscript straight a space straight c equals log presuperscript straight a straight b over straight c  log presuperscript 3 space 54 plus log presuperscript 3 space 2 minus log presuperscript 3 space 4 minus log presuperscript 3 space 9 equals log presuperscript 3 space fraction numerator 54.2 over denominator 9.4 end fraction equals log presuperscript 3 6 over 2 equals log presuperscript 3 space 3 equals 1 space left parenthesis straight A right parenthesis

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 c close parentheses end cell equals cell log presuperscript a space b plus log presuperscript a space c end cell row cell table attributes columnalign right columnspacing 2px end attributes row cell log presuperscript a space open parentheses b over c close parentheses end cell end table end cell equals cell table attributes columnalign left columnspacing 2px end attributes row cell log presuperscript a space b minus log presuperscript a space c end cell end table 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 open square brackets log space 200 close square brackets squared minus open square brackets log space 50 close square brackets squared dapat diperoleh sebagai berikut.

begin mathsize 12px style table attributes columnalign right center left columnspacing 2px end attributes row cell open square brackets log space 200 close square brackets squared minus open square brackets log space 50 close square brackets squared end cell equals cell open parentheses log space 200 plus log space 50 close parentheses open parentheses log space 200 minus log space 50 close parentheses end cell row blank equals cell open parentheses log space open parentheses 200 cross times 50 close parentheses close parentheses open parentheses log space open parentheses 200 over 50 close parentheses close parentheses end cell row blank equals cell open parentheses log space 10.000 close parentheses open parentheses log space 4 close parentheses end cell row blank equals cell open parentheses log space 10 to the power of 4 close parentheses open parentheses log space 4 close parentheses end cell row blank equals cell open parentheses 4 space log space 10 close parentheses open parentheses log space 4 close parentheses end cell row blank equals cell open parentheses 4 times 1 close parentheses log space 4 end cell row blank equals cell 4 space log space 4 end cell end table end style

Diperoleh bahwa open square brackets log space 200 close square brackets squared minus open square brackets log space 50 close square brackets squared equals 4 space log space 4.

Jadi, jawaban yang tepat adalah D.

0

Roboguru

Hasil dari  adalah ...

Pembahasan Soal:

Ingat sifat bentuk logaritma berikut.

alog(bc)=alog(b)+alog(c) 

  alog(cb)=alog(b)alog(c) 

alog(bm)=m(alogb)   

Penyelesaian soal di atas yaitu :

2log162log8+2log4======2log816×42log82log2332log2313   

Dengan demikian, hasil dari blank squared log 16 minus squared log 8 plus squared log 4 adalah 3.

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

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