Roboguru

Jika  dan  penyelesaian persamaan , nilai  adalah ....

Pertanyaan

Jika x subscript 1 dan x subscript 2 penyelesaian persamaan open parentheses 4 over 9 close parentheses to the power of x squared minus 3 end exponent cross times open parentheses 8 over 27 close parentheses to the power of 1 minus x end exponent equals 3 over 2, nilai open parentheses x subscript 1 minus x subscript 2 close parentheses squared adalah ....

  1. 9 over 4 

  2. 25 over 4 

  3. 49 over 4 

  4. 25 over 9 

  5. 25 

Pembahasan Soal:

Ingat kembali bentuk persamaan eksponen berikut:

straight a to the power of straight f open parentheses straight x close parentheses end exponent equals straight a to the power of straight m comma space straight a greater than 0 space rightwards arrow space straight f open parentheses straight x close parentheses equals straight m 

Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 4 over 9 close parentheses to the power of x squared minus 3 end exponent cross times open parentheses 8 over 27 close parentheses to the power of 1 minus x end exponent end cell equals cell 3 over 2 end cell row cell open parentheses 2 squared over 3 squared close parentheses to the power of x squared minus 3 end exponent cross times open parentheses 2 cubed over 3 cubed close parentheses to the power of 1 minus x end exponent end cell equals cell open parentheses 2 to the power of negative 1 end exponent over 3 to the power of negative 1 end exponent close parentheses end cell row cell open parentheses open parentheses 2 over 3 close parentheses squared close parentheses to the power of x squared minus 3 end exponent cross times open parentheses open parentheses 2 over 3 close parentheses cubed close parentheses to the power of 1 minus x end exponent end cell equals cell open parentheses 2 over 3 close parentheses to the power of negative 1 end exponent end cell row cell open parentheses 2 over 3 close parentheses to the power of 2 x squared minus 6 end exponent cross times open parentheses 2 over 3 close parentheses to the power of 3 minus 3 x end exponent end cell equals cell open parentheses 2 over 3 close parentheses to the power of negative 1 end exponent end cell row cell open parentheses 2 over 3 close parentheses to the power of open parentheses 2 x squared minus 6 close parentheses plus open parentheses 3 minus 3 x close parentheses end exponent end cell equals cell open parentheses 2 over 3 close parentheses to the power of negative 1 end exponent end cell row cell open parentheses 2 over 3 close parentheses to the power of 2 x squared minus 6 plus 3 minus 3 x end exponent end cell equals cell open parentheses 2 over 3 close parentheses to the power of negative 1 end exponent end cell row cell open parentheses 2 over 3 close parentheses to the power of 2 x squared minus 3 x minus 3 end exponent end cell equals cell open parentheses 2 over 3 close parentheses to the power of negative 1 end exponent end cell row cell 2 x squared minus 3 x minus 3 end cell equals cell negative 1 end cell row cell 2 x squared minus 3 x minus 3 plus 1 end cell equals 0 row cell 2 x squared minus 3 x minus 2 end cell equals 0 row cell open parentheses 2 x plus 1 close parentheses open parentheses x minus 2 close parentheses end cell equals 0 row x equals cell negative 1 half space atau space x equals 2 end cell end table 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses x subscript 1 minus x subscript 2 close parentheses squared end cell equals cell open parentheses negative 1 half minus 2 close parentheses squared end cell row blank equals cell open parentheses negative 1 half minus 4 over 2 close parentheses squared end cell row blank equals cell open parentheses negative 5 over 2 close parentheses squared end cell row blank equals cell 25 over 4 end cell end table 

Jadi, jawaban yang benar adalah B.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 19 Juli 2021

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