Tentukan penyelesaian persamaan berikut. b. 2log (x+7)+2log (x+6)+21log (x+10)=0

Pertanyaan

Tentukan penyelesaian persamaan berikut.

b. ${}^2\log\;\left(x+7\right)+{}^2\log\;\left(x+6\right)+{}^{\textstyle\frac12}\log\;\left(x+10\right)=0$

S. Amamah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

Ingat sifat bentuk logaritma:

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

Ingat ada persamaan logaritma berlaku jika log presuperscript a space f open parentheses x close parentheses equals log presuperscript a space g open parentheses x close parentheses comma space a greater than 0 dan a not equal to 1 maka dengan syarat dan .

Diketahui log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus log presuperscript begin inline style 1 half end style end presuperscript space open parentheses x plus 10 close parentheses equals 0 maka:

$begin mathsize 12px style table attributes columnalign right center left columnspacing 0px end attributes row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus log presuperscript begin inline style 1 half end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus log presuperscript begin inline style 2 to the power of negative 1 end exponent end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses plus open parentheses fraction numerator 1 over denominator negative 1 end fraction close parentheses log presuperscript begin inline style 2 end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses minus log presuperscript begin inline style 2 end style end presuperscript space open parentheses x plus 10 close parentheses end cell equals 0 row cell log presuperscript 2 space open parentheses x plus 7 close parentheses plus log presuperscript 2 space open parentheses x plus 6 close parentheses end cell equals cell log presuperscript begin inline style 2 end style end presuperscript space open parentheses x plus 10 close parentheses end cell row cell log presuperscript 2 space open parentheses x plus 7 close parentheses open parentheses x plus 6 close parentheses end cell equals cell log presuperscript 2 space open parentheses x plus 10 close parentheses end cell row cell log presuperscript 2 space open parentheses x squared plus 13 x plus 42 close parentheses end cell equals cell log presuperscript 2 space open parentheses x plus 10 close parentheses end cell row cell x squared plus 13 x plus 42 end cell equals cell x plus 10 end cell row cell x squared plus 13 x minus x plus 42 minus 10 end cell equals 0 row cell x squared plus 12 x plus 32 end cell equals 0 row cell open parentheses x plus 8 close parentheses open parentheses x plus 4 close parentheses end cell equals 0 end table end style$

diperoleh nilai yang memenuhi x equals negative 8 atau x equals negative 4

Syarat numerus 1)

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 7 end cell greater than 0 row x greater than cell negative 7 end cell end table

Syarat numerus 3)

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 6 end cell greater than 0 row x greater than cell negative 6 end cell end table

Syarat numerus 2)

table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 10 end cell greater than 0 row x greater than cell negative 10 end cell end table

Dari syarat numerus mengharuskan x greater than negative 6 maka nilai yang memenuhi adalah x equals negative 4 .

Dengan demikian himpunan penyelesaiany adalah open curly brackets negative 4 close curly brackets .