Roboguru

Penyelesaian persamaan  adalah ....

Pertanyaan

Penyelesaian persamaan open parentheses 1 over 16 close parentheses to the power of x minus 2 end exponent equals 64 adalah ....

  1. negative 7 over 2 

  2. negative 1 half 

  3. 1 half 

  4. 3 over 2 

  5. 7 over 2 

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 1 over 16 close parentheses to the power of x minus 2 end exponent end cell equals 64 row cell open parentheses 1 over 2 to the power of 4 close parentheses to the power of x minus 2 end exponent end cell equals cell 2 to the power of 6 end cell row cell open parentheses 2 to the power of negative 4 end exponent close parentheses to the power of x minus 2 end exponent end cell equals cell 2 to the power of 6 end cell row cell 2 to the power of negative 4 x plus 8 end exponent end cell equals cell 2 to the power of 6 end cell row cell negative 4 x plus 8 end cell equals 6 row cell negative 4 x end cell equals cell 6 minus 8 end cell row cell negative 4 x end cell equals cell negative 2 end cell row x equals cell fraction numerator negative 2 over denominator negative 4 end fraction end cell row x equals cell 1 half end cell end table 

Jadi, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 19 Juli 2021

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Produk Ruangguru

Produk Lainnya

Hubungi Kami

Ikuti Kami

©2021 Ruangguru. All Rights Reserved