Iklan

Iklan

Pertanyaan

( 9 1 − x 3 x + 2 ​ ​ ) 3 2 ​ = ....

 

  1. 3 

  2. 3 to the power of x 

  3. 3 to the power of x plus 1 end exponent 

  4. 3 to the power of 2 x plus 1 end exponent 

  5. 3 to the power of 2 x plus 2 end exponent 

Iklan

N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

jawaban yang benar adalah B.

Iklan

Pembahasan

Pembahasan
lock

Jika diberikanbilangan riil , danbilangan rasional maka berlaku sifatberikut. Sehingga dapat ditentukan hasil dari bentuk berikut. Sehingga diperoleh . Jadi, jawaban yang benar adalah B.

Jika diberikan bilangan riil a space dan space bb not equal to 0 dan bilangan rasional p comma space q comma space r maka berlaku sifat berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell a to the power of p over q end exponent end cell equals cell q-th root of a to the power of p end root end cell row cell b to the power of p over b to the power of q end cell equals cell b to the power of p minus q end exponent end cell row cell open parentheses a to the power of p close parentheses to the power of q end cell equals cell a to the power of p q end exponent end cell row cell open parentheses a to the power of p over b to the power of p close parentheses to the power of r end cell equals cell a to the power of p r end exponent over b to the power of q r end exponent end cell end table 

Sehingga dapat ditentukan hasil dari bentuk berikut. 

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses square root of 3 to the power of x plus 2 end exponent over 9 to the power of 1 minus x end exponent end root close parentheses to the power of 2 over 3 end exponent end cell equals cell open parentheses open parentheses 3 to the power of x plus 2 end exponent over 9 to the power of 1 minus x end exponent close parentheses to the power of 1 half end exponent close parentheses to the power of 2 over 3 end exponent end cell row blank equals cell 3 to the power of open parentheses x plus 2 close parentheses times begin display style 1 half end style times begin display style 2 over 3 end style end exponent over open parentheses 3 squared close parentheses to the power of open parentheses 1 minus x close parentheses times begin display style 1 half end style times begin display style 2 over 3 end style end exponent end cell row blank equals cell 3 to the power of begin display style fraction numerator x plus 2 over denominator 3 end fraction end style end exponent over 3 to the power of 2 times begin display style fraction numerator 1 minus x over denominator 3 end fraction end style end exponent end cell row blank equals cell 3 to the power of begin display style fraction numerator x plus 2 over denominator 3 end fraction end style end exponent over 3 to the power of begin display style fraction numerator 2 minus 2 x over denominator 3 end fraction end style end exponent end cell row blank equals cell 3 to the power of fraction numerator x plus 2 over denominator 3 end fraction minus fraction numerator 2 minus 2 x over denominator 3 end fraction end exponent end cell row blank equals cell 3 to the power of fraction numerator x plus 2 minus 2 plus 2 x over denominator 3 end fraction end exponent end cell row blank equals cell 3 to the power of fraction numerator 3 x over denominator 3 end fraction end exponent end cell row blank equals cell 3 to the power of x end cell end table  

Sehingga diperoleh open parentheses square root of 3 to the power of x plus 2 end exponent over 9 to the power of 1 minus x end exponent end root close parentheses to the power of 2 over 3 end exponent equals 3 to the power of x.

Jadi, jawaban yang benar adalah B.

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

169

Farhan Ramadhan

Pembahasan terpotong Makasih ❤️ Mudah dimengerti Pembahasan lengkap banget

Iklan

Iklan

Pertanyaan serupa

Sebutkan sifat-sifat eksponen.

3

4.0

Jawaban terverifikasi

RUANGGURU HQ

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

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

info@ruangguru.com

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia