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Akar-akar persamaan  adalah dan . Nilai

Pertanyaan

Akar-akar persamaan 9 to the power of x minus 6 times 3 to the power of x plus 8 equals 0 adalah x subscript 1 dan x subscript 2. Nilai x subscript 1 plus x subscript 2 space end subscript equals... space  

  1. log presuperscript 3 space 2

  2. log presuperscript 3 space 4

  3. log presuperscript 3 space 6

  4. log presuperscript 3 space 8

  5. log presuperscript 3 space 12

Pembahasan Soal:

Ingat kembali:

a to the power of b equals c rightwards arrow b equals log presuperscript a space c

log presuperscript a space c plus log presuperscript a space d equals log presuperscript a space open parentheses c times d close parentheses

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent comma space a not equal to 0

a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses equals g open parentheses x close parentheses comma space a not equal to 0

Misalkan:

3 to the power of x equals y 

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 9 to the power of x minus 6 times 3 to the power of x plus 8 end cell equals 0 row cell open parentheses 3 squared close parentheses to the power of x minus 6 times 3 to the power of x plus 8 end cell equals 0 row cell 3 to the power of 2 x end exponent minus 6 times 3 to the power of x plus 8 end cell equals 0 row cell open parentheses 3 to the power of x close parentheses squared minus 6 times 3 to the power of x plus 8 end cell equals 0 row cell y squared minus 6 y plus 8 end cell equals cell 0 space open parentheses faktorkan close parentheses end cell row cell open parentheses y minus 2 close parentheses open parentheses y minus 4 close parentheses end cell equals 0 row blank blank cell y equals 2 space atau space y equals 4 end cell end table   

Untuk y equals 2

   table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of x end cell equals y row cell 3 to the power of x end cell equals 2 row cell x subscript 1 end cell equals cell log presuperscript 3 space 2 end cell end table 

Untuk y equals 4

   table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of x end cell equals y row cell 3 to the power of x end cell equals 4 row cell x subscript 2 end cell equals cell scriptbase log space 4 end scriptbase presuperscript 3 end cell end table 

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 space end subscript end cell equals cell scriptbase log space 4 plus scriptbase log space 2 end scriptbase presuperscript 3 end scriptbase presuperscript 3 end cell row blank equals cell scriptbase log space 4 end scriptbase presuperscript 3 times 2 end cell row blank equals cell log presuperscript 3 space 8 end cell end table  

Dengan demikian, hasil dari x subscript 1 plus x subscript 2 space end subscript adalah  Error converting from MathML to accessible text. 

Jadi, jawaban yang tepat adalah D.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

B. Hary

Terakhir diupdate 13 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Akar-akar persamaan  adalah dan , dengan . Nilai

Pembahasan Soal:

Ingat kembali:

a to the power of b equals c rightwards arrow b equals log presuperscript a space c

log presuperscript a space c minus log presuperscript a space d equals log presuperscript a space open parentheses c over d close parentheses

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent comma space a not equal to 0

a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses equals g open parentheses x close parentheses comma space a not equal to 0

Misalkan:

5 to the power of x equals y 

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 25 to the power of x minus 9 times 5 to the power of x plus 18 end cell equals 0 row cell open parentheses 5 squared close parentheses to the power of x minus 9 times 5 to the power of x plus 18 end cell equals 0 row cell 5 to the power of 2 x end exponent minus 9 times 5 to the power of x plus 18 end cell equals 0 row cell open parentheses 5 to the power of x close parentheses squared minus 9 times 5 to the power of x plus 18 end cell equals 0 row cell y squared minus 9 y plus 18 end cell equals cell 0 space open parentheses difaktorkan close parentheses end cell row cell open parentheses y minus 3 close parentheses open parentheses y minus 6 close parentheses end cell equals cell 0 space end cell row blank blank cell y equals 3 space atau space y equals 6 end cell end table     

Untuk y equals 3

   table attributes columnalign right center left columnspacing 0px end attributes row cell 5 to the power of x end cell equals y row cell 5 to the power of x end cell equals 3 row x equals cell log presuperscript 5 space 3 end cell end table  

Untuk y equals 6

   table attributes columnalign right center left columnspacing 0px end attributes row cell 5 to the power of x end cell equals y row cell 5 to the power of x end cell equals 6 row x equals cell scriptbase log space 6 end scriptbase presuperscript 5 end cell end table 

Karena x subscript 1 greater than x subscript 2 maka x subscript 1 equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell blank presuperscript 5 end cell end table table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell log space 6 space dan space end cell end table x subscript 2 equals log presuperscript 5 space 3   Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 minus x subscript 2 space end subscript end cell equals cell scriptbase log space 6 plus scriptbase log space 3 end scriptbase presuperscript 5 end scriptbase presuperscript 5 end cell row blank equals cell scriptbase log space 6 over 3 end scriptbase presuperscript 5 end cell row blank equals cell log presuperscript 5 space 2 end cell end table  

Dengan demikian, hasil dari x subscript 1 plus x subscript 2 space end subscript adalah  Error converting from MathML to accessible text.

Jadi, jawaban yang tepat adalah A

0

Roboguru

Akar-akar persamaan  adalah dan . Nilai

Pembahasan Soal:

Ingat kembali:

a to the power of b equals c rightwards arrow b equals log presuperscript a space c

log presuperscript a space c plus log presuperscript a space d equals log presuperscript a space open parentheses c times d close parentheses 

table attributes columnalign right center left columnspacing 0px end attributes row cell Jika space a x squared plus b x plus c end cell equals cell 0 comma space maka space akar minus akarnya end cell row cell x subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell end table 

Misalkan:

3 to the power of x equals a 

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of 2 x plus 1 end exponent minus 21 times 3 to the power of x plus 15 end cell equals 0 row cell 3 to the power of 2 x end exponent times 3 minus 21 times 3 to the power of x plus 15 end cell equals 0 row cell 3 times open parentheses 3 to the power of x close parentheses squared minus 21 times 3 to the power of x plus 15 end cell equals 0 row cell 3 times a squared minus 21 times a plus 15 end cell equals cell 0 space open parentheses dibagi space 3 close parentheses end cell row cell a squared minus 7 a plus 5 end cell equals 0 row blank blank cell a equals 1 comma space b equals negative 7 comma space c equals 5 end cell end table       

Dengan menggunakan rumus ABC, maka kita tentukan akar-akar dari persamaan di atas:

table attributes columnalign right center left columnspacing 0px end attributes row cell a subscript 1 comma 2 end subscript end cell equals cell fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction end cell row blank equals cell fraction numerator negative open parentheses negative 7 close parentheses plus-or-minus square root of open parentheses negative 7 close parentheses squared minus 4 open parentheses 1 close parentheses open parentheses 5 close parentheses end root over denominator 2 open parentheses 1 close parentheses end fraction end cell row blank equals cell fraction numerator 7 plus-or-minus square root of 49 minus 20 end root over denominator 2 end fraction end cell row blank equals cell fraction numerator 7 plus-or-minus square root of 29 over denominator 2 end fraction end cell row blank blank blank row cell a subscript 1 end cell equals cell fraction numerator 7 plus square root of 29 over denominator 2 end fraction end cell row cell a subscript 2 end cell equals cell fraction numerator 7 minus square root of 29 over denominator 2 end fraction end cell end table 

Sehingga diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of x end cell equals cell a subscript 1 end cell row cell 3 to the power of x end cell equals cell fraction numerator 7 plus square root of 29 over denominator 2 end fraction end cell row cell x subscript 1 end cell equals cell log presuperscript 3 space fraction numerator 7 plus square root of 29 over denominator 2 end fraction end cell row blank blank blank row cell 3 to the power of x end cell equals cell a subscript 2 end cell row cell 3 to the power of x end cell equals cell fraction numerator 7 minus square root of 29 over denominator 2 end fraction space end cell row cell x subscript 2 end cell equals cell log presuperscript 3 space fraction numerator 7 minus square root of 29 over denominator 2 end fraction end cell end table  

Sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 plus x subscript 2 end cell equals cell log presuperscript 3 space fraction numerator 7 plus square root of 29 over denominator 2 end fraction plus log presuperscript 3 space fraction numerator 7 minus square root of 29 over denominator 2 end fraction end cell row blank equals cell log presuperscript 3 space open parentheses fraction numerator 7 plus square root of 29 over denominator 2 end fraction times fraction numerator 7 minus square root of 29 over denominator 2 end fraction close parentheses end cell row blank equals cell log presuperscript 3 space open parentheses fraction numerator 49 minus 29 over denominator 4 end fraction close parentheses end cell row blank equals cell log presuperscript 3 space 20 over 4 end cell row blank equals cell log presuperscript 3 space 5 end cell end table

Dengan demikian, hasil dari x subscript 1 plus x subscript 2 space end subscript adalah  log presuperscript 3 space 5

Jadi, jawaban yang tepat adalah D

0

Roboguru

Nilai x yang memenuhi persamaan

Pembahasan Soal:

Ingat kembali:

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent comma space a not equal to 0 

a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses equals g open parentheses x close parentheses

Misalkan:

3 to the power of x equals y

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 9 to the power of x minus 7 times 3 to the power of x minus 18 end cell equals 0 row cell open parentheses 3 squared close parentheses to the power of x minus 7 times 3 to the power of x minus 18 end cell equals 0 row cell 3 to the power of 2 x end exponent minus 7 times 3 to the power of x minus 18 end cell equals 0 row cell open parentheses 3 to the power of x close parentheses squared minus 7 times 3 to the power of x minus 18 end cell equals 0 row cell y squared minus 7 y minus 18 end cell equals cell 0 space end cell row cell open parentheses y minus 9 close parentheses open parentheses y plus 2 close parentheses end cell equals cell 0 space space open parentheses dengan space cara space pemfaktoran close parentheses end cell row blank blank cell y equals 9 space atau space y equals negative 2 end cell end table  

Untuk y equals negative 2  tidak memenuhi persamaan di atas, karena a bernilai negatif, hanya y equals 9 yang memenuhi, sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of x end cell equals y row cell 3 to the power of x end cell equals 9 row cell 3 to the power of x end cell equals cell 3 squared end cell row x equals 2 end table

Dengan demikian, Nilai x yang memenuhi persamaan 9 to the power of x minus 7 times 3 to the power of x minus 18 equals 0  adalah 2

Jadi, jawaban yang tepat adalah B

0

Roboguru

Akar-akar persamaan  adalah dan . Nilai

Pembahasan Soal:

Ingat kembali:

a to the power of n equals stack stack a cross times a cross times a cross times... cross times a with underbrace below with a space s e b a n y a k space n below comma space a not equal to 0

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent comma space a not equal to 0

1 over a to the power of n equals a to the power of negative n end exponent comma space a not equal to 0

a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses equals g open parentheses x close parentheses comma space a not equal to 0

Misalkan:

4 to the power of x equals y 

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of x minus 12 plus 32 times 4 to the power of negative x end exponent end cell equals 0 row cell 4 to the power of x minus 12 plus 32 over 4 to the power of x end cell equals 0 row cell y minus 12 plus 32 over y end cell equals cell 0 space open parentheses sama minus sama space dikali space y close parentheses end cell row cell y squared minus 12 y plus 32 end cell equals cell 0 space open parentheses difaktorkan close parentheses end cell row cell open parentheses y minus 8 close parentheses open parentheses y minus 4 close parentheses end cell equals cell 0 space end cell row blank blank cell y equals 8 space atau space y equals 4 end cell end table  

Untuk y equals 8

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of x end cell equals y row cell 4 to the power of x end cell equals 8 row cell open parentheses 2 squared close parentheses to the power of x end cell equals cell 2 cubed end cell row cell 2 to the power of 2 x end exponent end cell equals cell 2 cubed end cell row cell 2 x end cell equals 3 row cell x subscript 1 end cell equals cell 3 over 2 end cell end table   

Untuk y equals 4

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of x end cell equals y row cell 4 to the power of x end cell equals 4 row cell open parentheses 2 squared close parentheses to the power of x end cell equals cell 2 squared end cell row cell 2 to the power of 2 x end exponent end cell equals cell 2 squared end cell row cell 2 x end cell equals 2 row x equals cell 2 over 2 end cell row cell x subscript 2 end cell equals 1 end table   

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell x subscript 1 times x subscript 2 space end subscript end cell equals cell 3 over 2 times 1 end cell row blank equals cell 3 over 2 end cell row blank equals cell 1 1 half end cell end table  

Dengan demikian, hasil dari x subscript 1 times x subscript 2 space end subscript adalah  table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell 1 1 half end cell end table 

Jadi, jawaban yang tepat adalah C

0

Roboguru

Nilai x yang memenuhi persamaan  adalah...

Pembahasan Soal:

Ingat kembali:

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent comma space a not equal to 0

a to the power of f open parentheses x close parentheses end exponent equals a to the power of g open parentheses x close parentheses end exponent rightwards arrow f open parentheses x close parentheses equals g open parentheses x close parentheses comma space a not equal to 0 

Misalkan:

2 to the power of 2 x end exponent equals y 

Sehingga diperoleh perhitungan:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of 4 x end exponent minus 7 times 2 to the power of 2 x end exponent minus 8 end cell equals 0 row cell 2 to the power of 2 times 2 x end exponent minus 7 times 2 to the power of 2 x end exponent minus 8 end cell equals 0 row cell open parentheses 2 to the power of 2 x end exponent close parentheses squared minus 7 times 2 to the power of 2 x end exponent minus 8 end cell equals 0 row cell y squared minus 7 y minus 8 end cell equals 0 row cell open parentheses y minus 8 close parentheses open parentheses y plus 1 close parentheses end cell equals cell 0 space open parentheses mengunakan space pemfaktoran close parentheses end cell row blank blank cell y equals 8 space atau space y equals negative 1 end cell end table

Untuk y equals negative 1 tidak memenuhi fungsi di atas, karena a negatif, maka hanya y equals 8 yang memenuhi, sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of 2 x end exponent end cell equals y row cell 2 to the power of 2 x end exponent end cell equals 8 row cell 2 to the power of 2 x end exponent end cell equals cell 2 cubed end cell row cell 2 x end cell equals 3 row x equals cell 3 over 2 end cell end table

Dengan demikian, Nilai x yang memenuhi persamaan 2 to the power of 4 x end exponent minus 7 times 2 to the power of 2 x end exponent minus 8 equals 0 adalah  3 over 2

Jadi, jawaban yang tepat adalah C

0

Roboguru

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