Pertanyaan

Akar-akar persamaan 3 2 x + 1 − 21 ⋅ 3 x + 15 = 0 adalah x 1 dan x 2 . Nilai x 1 + x 2 = ...

Akar-akar persamaan $3^{2 x + 1} - 21 \cdot 3^{x} + 15 = 0$ adalah $x_{1}$ dan $x_{2}$ . Nilai $x_{1} + x_{2} = ...$

M. Nasrullah

Master Teacher

Mahasiswa/Alumni Universitas Negeri Makassar

Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah D

Pembahasan

Ingat kembali: Misalkan: Sehingga diperoleh perhitungan: Dengan menggunakan rumus ABC, maka kita tentukan akar-akar dari persamaan di atas: Sehingga diperoleh: Sehingga Dengan demikian, hasil dari adalah Jadi, jawaban yang tepat adalah D

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:

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