Roboguru

Jika  dan  maka

Pertanyaan

Jika x not equal to 0 dan y not equal to 0 maka open square brackets open parentheses x. y to the power of a close parentheses to the power of 1 minus a end exponent close square brackets squared. open square brackets open parentheses x. y to the power of a close parentheses to the power of a minus 1 end exponent close square brackets squared equals... 

  1. open parentheses x y close parentheses to the power of 2 a minus 2 a squared end exponent 

  2. open parentheses x y close parentheses to the power of 4 a minus 4 a squared end exponent 

  3. x y 

  4. x squared y to the power of 2 a squared end exponent 

  5. 1 

Pembahasan Video:

Pembahasan Soal:

Ingat!

Sifat bilangan berpangkat:

  • a to the power of m cross times a to the power of n equals a to the power of m plus n end exponent 
  • open parentheses a to the power of m close parentheses to the power of n equals a to the power of m cross times n end exponent 
  • a to the power of 0 equals 1 comma space a not equal to 0 

Maka:

[(x.ya)1a]2.[(x.ya)a1]2=========[x1a.(ya)1a]2.[(xa1.(ya)a1)]2[x1aya×(1a)]2[(xa1ya×(a1))]2[x1ayaa2]2[xa1ya2a]2x(1a)×2y(aa2)×2x(a1)×2y(a2a)×2x22ay2a2a2x2a2y2a22ax22a+2a2y2a2a2+2a22ax0y0(1)(1)1


Jadi, jawaban yang tepat adalah E.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

N. Puspita

Terakhir diupdate 13 September 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

Pertanyaan yang serupa

Jumlah akar-akar persamaan  adalah ....

Pembahasan Soal:

Ingat kembali:

  • a to the power of m plus n end exponent equals a to the power of m times a to the power of n
  • open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent 
  • Bentuk a times p to the power of 2 x end exponent plus b times p to the power of x plus c equals 0 merupakan persamaan eksponen yang dapat diselesaikan dengan memisalkan y equals p to the power of x sehingga diperoleh bentuk persamaan kuadrat a times y squared plus b times y plus c equals 0. Selanjutnya, penyelesaian dapat diperoleh dengan menentukan terlebih dahulu akar-akar persamaan kuadrat yang terbentuk.
  • Jika diketahui persamaan kuadrat a times y squared plus b times y plus c equals 0, maka hasil kali akar-akar persamaan kuadrat adalah table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 times y subscript 2 end cell equals cell c over a end cell end table.
  • Jika af(x)=ap dengan a>0 dan a=1, maka f open parentheses x close parentheses equals p

Oleh karena itu, dengan memisalkan y equals 3 to the power of x, maka diperoleh

table attributes columnalign right center left columnspacing 0px end attributes row cell 3 times 9 to the power of x minus 10 times 3 to the power of x plus 3 end cell equals 0 row cell 3 times open parentheses 3 squared close parentheses to the power of x minus 10 times 3 to the power of x plus 3 end cell equals 0 row cell 3 times 3 to the power of 2 x end exponent minus 10 times 3 to the power of x plus 3 end cell equals 0 row cell 3 times y squared minus 10 times y plus 3 end cell equals cell 0 space space space space space midline horizontal ellipsis open parentheses 1 close parentheses end cell end table

Dari persamaan open parentheses 1 close parentheses, diperoleh hasil kali akar-akarnya adalah 

table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 times y subscript 2 end cell equals cell 3 over 3 end cell row blank equals 1 end table

Dengan mengembalikan ke bentuk y equals 3 to the power of x, diperoleh

y1y23x13x23x1+x23x1+x2====11130

sehingga karena a=3 yang berarti a>0 dan a=1, maka diperoleh x1+x2=0.

Jadi, jumlah akar-akar persamaan 3 times 9 to the power of x minus 10 times 3 to the power of x plus 3 equals 0adalah 0.

Oleh karena itu, jawaban yang benar adalah C.space 

0

Roboguru

Jika  tentukanlah nilai dari .

Pembahasan Soal:

Sifat bilangan berpangkat:

table attributes columnalign right center left columnspacing 2px end attributes row cell a to the power of m cross times a to the power of n end cell equals cell a to the power of m plus n end exponent end cell row cell open parentheses a to the power of m close parentheses to the power of n end cell equals cell a to the power of m cross times n end exponent end cell row cell a to the power of 0 end cell equals cell 1 comma space a not equal to 0 end cell end table

Dengan mengkuadratkan kedua ruas dan menggunakan sifat-sifat bilangan berpangkat di atas, dapat diperoleh perhitungan sebagai berikut.

table attributes columnalign right center left columnspacing 2px end attributes row cell 2 to the power of x plus 2 to the power of negative x end exponent end cell equals cell 5 space end cell row cell open parentheses 2 to the power of x plus 2 to the power of negative x end exponent close parentheses squared end cell equals cell 5 squared end cell row cell open parentheses 2 to the power of x plus 2 to the power of negative x end exponent close parentheses open parentheses 2 to the power of x plus 2 to the power of negative x end exponent close parentheses end cell equals 25 row cell 2 to the power of x times 2 to the power of x plus 2 to the power of x times 2 to the power of negative x end exponent plus 2 to the power of negative x end exponent times 2 to the power of x plus 2 to the power of negative x end exponent times 2 to the power of negative x end exponent end cell equals 25 row cell 2 to the power of x plus x end exponent plus 2 to the power of x minus x end exponent plus 2 to the power of negative x plus x end exponent plus 2 to the power of negative x minus x end exponent end cell equals 25 row cell 2 to the power of 2 x end exponent plus 2 to the power of 0 plus 2 to the power of 0 plus 2 to the power of negative 2 x end exponent end cell equals 25 row cell open parentheses 2 squared close parentheses to the power of x plus 1 plus 1 plus open parentheses 2 squared close parentheses to the power of negative x end exponent end cell equals 25 row cell 4 to the power of x plus 2 plus 4 to the power of negative x end exponent end cell equals 25 row cell 4 to the power of x plus 4 to the power of negative x end exponent end cell equals cell 25 minus 2 end cell row cell 4 to the power of x plus 4 to the power of negative x end exponent end cell equals 23 end table

Jadi, nilai dari 4 to the power of x plus 4 to the power of negative x end exponent adalah 23.

2

Roboguru

Sederhanakan. e. (42⋅2−4)⋅(52⋅3−3)2

Pembahasan Soal:

Jika a bilangan real dan n bilangan bulat positif, maka pangkat n dari a ditulis an didefinisikan sebagai berikut.

an=aaaaaasebanyakn

Ingat sifat bilangan berpangkat (eksponen) berikut.

aman=am+n,a=0

(ab)m=ambm

(am)n=amn

am=am1,a=0

a0=1,a=0

Penyelesaian soal di atas adalah sebagai berikut.

(4224)(5233)2=======((22)224)(52)2(33)2(22224)(5436)(2424)(5436)(24+(4))543620(54361)1(54361)3654

Dengan demikian, bentuk sederhana dari (4224)(5233)2=3654

0

Roboguru

Tentukan hasil pengerjaan hitung berikut. a.

Pembahasan Soal:

Sifat-sifat bilangan berpangkat di antaranya sebagai berikut.

a to the power of m times a to the power of n equals a to the power of m plus n end exponent

a to the power of m over a to the power of n equals a to the power of m minus n end exponent,  a not equal to 0

a to the power of 0 equals 1a not equal to 0

open parentheses a times b close parentheses to the power of m equals a to the power of m times b to the power of m

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m n end exponent

Penyelesaian soal di atas adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 2 to the power of negative 3 end exponent cross times 9 to the power of negative 2 end exponent over denominator 18 to the power of negative 3 end exponent end fraction plus fraction numerator 6 to the power of 12 cross times 24 to the power of negative 2 end exponent over denominator 12 cubed end fraction end cell row blank equals cell fraction numerator 2 to the power of negative 3 end exponent cross times open parentheses 3 squared close parentheses to the power of negative 2 end exponent over denominator open parentheses 2 cross times 3 squared close parentheses to the power of negative 3 end exponent end fraction plus fraction numerator open parentheses 2 cross times 3 close parentheses to the power of 12 cross times open parentheses 2 cubed cross times 3 close parentheses to the power of negative 2 end exponent over denominator open parentheses 2 squared cross times 3 close parentheses cubed end fraction end cell row blank equals cell fraction numerator 2 to the power of negative 3 end exponent cross times 3 to the power of negative 4 end exponent over denominator 2 to the power of negative 3 end exponent cross times 3 to the power of negative 6 end exponent end fraction plus fraction numerator 2 to the power of 12 cross times 3 to the power of 12 cross times 2 to the power of negative 6 end exponent cross times 3 to the power of negative 2 end exponent over denominator 2 to the power of 6 cross times 3 cubed end fraction end cell row blank equals cell open parentheses 2 to the power of 0 cross times 3 to the power of negative 4 minus open parentheses negative 6 close parentheses end exponent close parentheses plus fraction numerator 2 to the power of 12 plus open parentheses negative 6 close parentheses end exponent cross times 3 to the power of 12 plus open parentheses negative 2 close parentheses end exponent over denominator 2 to the power of 6 cross times 3 cubed end fraction end cell row blank equals cell open parentheses 1 cross times 3 squared close parentheses plus fraction numerator 2 to the power of 6 cross times 3 to the power of 10 over denominator 2 to the power of 6 cross times 3 cubed end fraction end cell row blank equals cell 3 squared plus open parentheses 2 to the power of 6 minus 6 end exponent cross times 3 to the power of 10 minus 3 end exponent close parentheses end cell row blank equals cell 3 squared plus open parentheses 2 to the power of 0 cross times 3 to the power of 7 close parentheses end cell row blank equals cell 3 squared plus open parentheses 1 cross times 3 to the power of 7 close parentheses end cell row blank equals cell 3 squared plus 3 to the power of 7 end cell row blank equals cell 9 plus 2.187 end cell row blank equals cell 2.196 end cell end table

Dengan demikian, hasil dari pengerjaan hitung Error converting from MathML to accessible text.

10

Roboguru

Diketahui , , dan . Nilai  adalah ...

Pembahasan Soal:

Sifat-sifat bilangan berpangkat di antaranya sebagai berikut.

a to the power of m times a to the power of n equals a to the power of m plus n end exponent

a to the power of m over a to the power of n equals a to the power of m minus n end exponent,  a not equal to 0

a to the power of negative m end exponent equals 1 over a to the power of m,  a not equal to 0

open parentheses a to the power of m close parentheses to the power of n equals a to the power of m times n end exponent

a to the power of 0 equals 1a not equal to 0

Penyelesaian soal di atas adalah sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses a to the power of negative 1 end exponent close parentheses squared cross times b to the power of 4 over c to the power of negative 3 end exponent end cell equals cell open parentheses 4 to the power of negative 1 end exponent close parentheses squared cross times 2 to the power of 4 over open parentheses begin display style 1 half end style close parentheses to the power of negative 3 end exponent end cell row blank equals cell 4 to the power of negative 2 end exponent cross times 2 to the power of 4 over open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of negative 3 end exponent end cell row blank equals cell open parentheses 2 squared close parentheses to the power of negative 2 end exponent cross times 2 to the power of 4 over 2 cubed end cell row blank equals cell 2 to the power of negative 4 end exponent cross times 2 to the power of 4 over 2 cubed end cell row blank equals cell 2 to the power of negative 4 plus 4 end exponent over 2 cubed end cell row blank equals cell 2 to the power of 0 over 2 cubed end cell row blank equals cell 1 over 8 end cell end table

Oleh karena itu, jawaban yang tepat adalah C.

1

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

Tapi Roboguru masih mau belajar. Menurut kamu pembahasan kali ini sudah membantu, belum?

Membantu

Kurang Membantu

Apakah pembahasan ini membantu?

Belum menemukan yang kamu cari?

Post pertanyaanmu ke Tanya Jawab, yuk

Mau Bertanya

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