Roboguru

Nilai x yang memenuhi pertidaksamaan 32x+1−63×9x−1+108≥0 adalah ...

Pertanyaan

Nilai x yang memenuhi pertidaksamaan 3 to the power of 2 x plus 1 end exponent minus 63 cross times 9 to the power of x minus 1 end exponent plus 108 greater or equal than 0 adalah ...

  1. x greater or equal than 3

  2. x greater or equal than 2

  3. x less or equal than 3 over 2

  4. x greater or equal than 3 over 2

  5. x less or equal than 2 over 3

Pembahasan Soal:

Pertama, gunakan sifat-sifat bilangan berpangkat untuk menyederhanakan pertidaksamaan tersebut.


table attributes columnalign right center left columnspacing 0px end attributes row cell 3 to the power of 2 straight x plus 1 end exponent minus 63 cross times 9 to the power of straight x minus 1 end exponent plus 108 end cell greater or equal than 0 row cell 3 to the power of 1 cross times 3 to the power of 2 x end exponent minus 63 cross times open parentheses 3 squared close parentheses to the power of x minus 1 end exponent plus 108 end cell greater or equal than 0 row cell 3 cross times 3 to the power of 2 x end exponent minus 63 cross times 3 to the power of negative 2 end exponent cross times 3 to the power of 2 x end exponent plus 108 end cell greater or equal than 0 row cell 3 cross times 3 to the power of 2 x end exponent minus 63 over 9 cross times 3 to the power of 2 x end exponent plus 108 end cell greater or equal than 0 row cell 3 cross times 3 to the power of 2 x end exponent minus 7 cross times 3 to the power of 2 x end exponent plus 108 end cell greater or equal than 0 row cell negative 4 cross times 3 to the power of 2 x end exponent plus 108 end cell greater or equal than 0 row cell negative 4 cross times 3 to the power of 2 x end exponent end cell greater or equal than cell negative 108 end cell row cell 3 to the power of 2 x end exponent end cell less or equal than 27 row cell 3 to the power of 2 x end exponent end cell less or equal than cell 3 cubed end cell end table


Pertidaksamaan tersebut adalah pertidaksamaan eksponen dengan basis a greater than 1, mempunyai penyelesaian:


table attributes columnalign right center left columnspacing 0px end attributes row cell 2 x end cell less or equal than 3 row x less or equal than cell 3 over 2 end cell end table


Jadi, himpunan penyelesaiannya adalah HP equals open curly brackets x vertical line x less or equal than 3 over 2 close curly brackets.

Oleh karena itu, jawaban yang benar adalah C.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

A. Salim

Mahasiswa/Alumni Universitas Pelita Harapan

Terakhir diupdate 06 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Pertanyaan yang serupa

Penyelesaian pertidaksamaan 4⋅2x2+1>(21​)5x+1 adalah ...

Pembahasan Soal:

Diketahui:

Pertidaksamaan 4 times 2 to the power of x squared plus 1 end exponent greater than open parentheses 1 half close parentheses to the power of 5 x plus 1 end exponent

Ditanya:

Pernyelesaian pertidaksamaan tersebut.

Jika terdapat suatu pertidaksamaan dimana a greater than 1 dan a to the power of f open parentheses x close parentheses end exponent greater than a to the power of g left parenthesis x right parenthesis end exponent maka f open parentheses x close parentheses greater than g open parentheses x close parentheses.

Perlu diingat bahwa:

1 over k to the power of l equals k to the power of negative l end exponent k to the power of open parentheses l plus m close parentheses end exponent equals k to the power of l times k to the power of m open parentheses k to the power of l close parentheses to the power of m equals k to the power of l times m end exponent 

Perhatikan perhitungan berikut:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 times 2 to the power of x squared plus 1 end exponent end cell greater than cell open parentheses 1 half close parentheses to the power of 5 x plus 1 end exponent end cell row cell 2 squared times 2 to the power of x squared plus 1 end exponent end cell greater than cell open parentheses 2 to the power of negative 1 end exponent close parentheses to the power of 5 x plus 1 end exponent end cell row cell 2 to the power of x squared plus 1 plus 2 end exponent end cell greater than cell 2 to the power of negative 5 x minus 1 end exponent end cell row cell 2 to the power of x squared plus 3 end exponent end cell greater than cell 2 to the power of negative 5 x minus 1 end exponent end cell row cell x squared plus 3 end cell greater than cell negative 5 x minus 1 end cell row cell x squared plus 3 plus 5 x plus 1 end cell greater than cell negative 5 x plus 5 x minus 1 plus 1 end cell row cell x squared plus 5 x plus 4 end cell greater than 0 row cell open parentheses x plus 1 close parentheses open parentheses x plus 4 close parentheses end cell greater than 0 end table
table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 1 end cell equals 0 row x equals cell negative 1 end cell end table atau  table attributes columnalign right center left columnspacing 0px end attributes row cell x plus 4 end cell equals 0 row x equals cell negative 4 end cell end table        

Dengan menggunakan garis bilangan dan melakukan uji titik pada interval, diperoleh solusi dari x squared plus 5 x plus 4 greater than 0 adalah

Oleh sebab itu, didapat himpunan penyelesaiannya adalah open curly brackets x space left enclose space x less than negative 4 end enclose space atau space x greater than negative 1 close curly brackets.

Sehingga, penyelesaian pertidaksamaan 4 times 2 to the power of x squared plus 1 end exponent greater than open parentheses 1 half close parentheses to the power of 5 x plus 1 end exponent adalah x less than negative 4 space atau space x greater than negative 1.

Jadi, jawaban yang tepat adalah B.

3

Roboguru

Tentukanlah interval di mana grafik fungsi y=24x+2​ berada di bawah grafik dari fungsi y=23x−11​​.

Pembahasan Soal:

Diketahui:

Grafik fungsi y equals 2 square root of 4 to the power of x plus 2 end exponent end root berada di bawah grafik  fungsi y equals square root of 1 over 2 to the power of 3 x minus 1 end exponent end root.

Ditanya:

Interval grafik tersebut

Grafik fungsi y subscript 1 equals 2 square root of 4 to the power of x plus 2 end exponent end root berada di bawah grafik fungsi y subscript 2 equals square root of 1 over 2 to the power of 3 x minus 1 end exponent end root apabila y subscript 1 less than y subscript 2.

table attributes columnalign right center left columnspacing 0px end attributes row cell y subscript 1 end cell less than cell y subscript 2 end cell row cell 2 square root of 4 to the power of x plus 2 end exponent end root end cell less than cell square root of 1 over 2 to the power of 3 x minus 1 end exponent end root end cell row cell 2 open parentheses 4 to the power of x plus 2 end exponent close parentheses to the power of begin inline style 1 half end style end exponent end cell less than cell open parentheses 1 over 2 to the power of 3 x minus 1 end exponent close parentheses to the power of begin inline style 1 half end style end exponent end cell row cell 2 open parentheses open parentheses 2 squared close parentheses to the power of x plus 2 end exponent close parentheses to the power of begin inline style 1 half end style end exponent end cell less than cell open parentheses 2 to the power of negative open parentheses 3 x minus 1 close parentheses end exponent close parentheses to the power of begin inline style 1 half end style end exponent end cell row cell 2 open parentheses 2 to the power of 2 x plus 4 end exponent close parentheses to the power of begin inline style 1 half end style end exponent end cell less than cell open parentheses 2 to the power of negative 3 x plus 1 end exponent close parentheses to the power of begin inline style 1 half end style end exponent end cell row cell 2 open parentheses 2 to the power of x plus 2 end exponent close parentheses end cell less than cell open parentheses 2 to the power of begin inline style fraction numerator negative 3 x plus 1 over denominator 2 end fraction end style end exponent close parentheses end cell row cell 2 to the power of 1 open parentheses 2 to the power of x plus 2 end exponent close parentheses end cell less than cell open parentheses 2 to the power of begin inline style fraction numerator negative 3 x plus 1 over denominator 2 end fraction end style end exponent close parentheses end cell row cell open parentheses 2 to the power of x plus 2 plus 1 end exponent close parentheses end cell less than cell open parentheses 2 to the power of begin inline style fraction numerator negative 3 x plus 1 over denominator 2 end fraction end style end exponent close parentheses end cell row cell open parentheses 2 to the power of x plus 3 end exponent close parentheses end cell less than cell open parentheses 2 to the power of begin inline style fraction numerator negative 3 x plus 1 over denominator 2 end fraction end style end exponent close parentheses end cell end table

Karena basis pertidaksamaan di atas lebih dari 1, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 to the power of x plus 3 end exponent close parentheses end cell less than cell open parentheses 2 to the power of begin inline style fraction numerator negative 3 x plus 1 over denominator 2 end fraction end style end exponent close parentheses end cell row cell x plus 3 end cell less than cell negative fraction numerator 3 x plus 1 over denominator 2 end fraction end cell row cell 2 x plus 6 end cell less than cell negative 3 x plus 1 end cell row cell 2 x plus 3 x end cell less than cell 1 minus 6 end cell row cell 5 x end cell less than cell negative 5 end cell row x less than cell negative 1 end cell end table

Jadi, interval fungsi tersebut adalah x less than negative 1.

1

Roboguru

Nilai x yang memenuhi pertidaksamaan 41−x−5⋅22−x+16<0 adalah ....

Pembahasan Soal:

Ingat kembali bentuk persamaan eksponen berikut:

begin mathsize 12px style A open parentheses a to the power of f open parentheses x close parentheses end exponent close parentheses squared plus B open parentheses a to the power of f open parentheses x close parentheses end exponent close parentheses plus C equals 0 comma space a greater than 0 comma space a not equal to 1 comma space A not equal to 0 comma space dan space A comma B comma space C space element of R end style 

Untuk menyelesaikan bentuk persamaan ini digunakan pemisalan y equals a to the power of f open parentheses x close parentheses end exponent sehingga diperoleh A y squared plus B y plus C equals 0. Setelah nilai y diperoleh, substitusikan kembali pada pemisalan y equals a to the power of f open parentheses x close parentheses end exponent sehingga diperoleh nilai x.

Diketahui pertidaksamaan 4 to the power of 1 minus x end exponent minus 5 times 2 to the power of 2 minus x end exponent plus 16 less than 0. Maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of 1 minus x end exponent minus 5 times 2 to the power of 2 minus x end exponent plus 16 end cell less than 0 row cell 4 to the power of 1 times 4 to the power of negative x end exponent minus 5 times 2 squared times 2 to the power of negative x end exponent plus 16 end cell less than 0 row cell 4 times open parentheses 2 squared close parentheses to the power of negative x end exponent minus 5 times 4 times 2 to the power of negative x end exponent plus 16 end cell less than 0 row cell 4 times open parentheses 2 to the power of negative x end exponent close parentheses squared minus 20 times 2 to the power of negative x end exponent plus 16 end cell less than 0 end table 

Misalkan 2 to the power of negative x end exponent equals y, berarti:

table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of 1 minus x end exponent minus 5 times 2 to the power of 2 minus x end exponent plus 16 end cell less than 0 row cell 4 to the power of 1 times 4 to the power of negative x end exponent minus 5 times 2 squared times 2 to the power of negative x end exponent plus 16 end cell less than 0 row cell 4 times open parentheses 2 squared close parentheses to the power of negative x end exponent minus 5 times 4 times 2 to the power of negative x end exponent plus 16 end cell less than 0 row cell 4 times open parentheses 2 to the power of negative x end exponent close parentheses squared minus 20 times 2 to the power of negative x end exponent plus 16 end cell less than 0 row cell 4 times y squared minus 20 times y plus 16 end cell less than 0 row cell 4 y squared minus 20 y plus 16 end cell less than 0 row cell y squared minus 5 y plus 4 end cell less than 0 row cell open parentheses y minus 1 close parentheses open parentheses y minus 4 close parentheses end cell less than 0 row y equals cell 1 space atau space y equals 4 end cell end table 

Sehingga:

table attributes columnalign right center left columnspacing 0px end attributes row y equals 1 row cell 2 to the power of negative x end exponent end cell equals 1 row cell 2 to the power of negative x end exponent end cell equals cell 2 to the power of 0 end cell row cell negative x end cell equals 0 row x equals 0 end table     atau     table attributes columnalign right center left columnspacing 0px end attributes row y equals 4 row cell 2 to the power of negative x end exponent end cell equals 4 row cell 2 to the power of negative x end exponent end cell equals cell 2 squared end cell row cell negative x end cell equals 2 row x equals cell negative 2 end cell end table 

Uji titik pada 41x522x+16<0

misalkan ambil titik x=1

41x522x+16===41(1)522(1)+161658+168()

Daerah penyelesaiannya dapat digambarkan sebagai berikut:

Dari gambar di atas dapat diketahui nilai x yang memenuhi negative 2 less than x less than 0.

Jadi, jawaban yang benar adalah E.

0

Roboguru

Tentukan nilai x yang memenuhi pertidaksamaan 4x2−x−2×2x2−5x+4&lt;161​.

Pembahasan Soal:

Pertama kita gunakan sifat-sifat bilangan berpangkat untuk menyederhanakan pertidaksamaan tersebut.


table attributes columnalign right center left columnspacing 0px end attributes row cell 4 to the power of straight x squared minus straight x minus 2 end exponent cross times 2 to the power of straight x squared minus 5 straight x plus 4 end exponent end cell less than cell 1 over 16 end cell row cell open parentheses 2 squared close parentheses to the power of blank to the power of straight x squared minus straight x minus 2 end exponent end exponent cross times 2 to the power of straight x squared minus 5 straight x plus 4 end exponent end cell less than cell 2 to the power of negative 4 end exponent end cell row cell 2 to the power of 2 straight x squared minus 2 straight x minus 4 end exponent cross times 2 to the power of straight x squared minus 5 straight x plus 4 end exponent end cell less than cell 2 to the power of negative 4 end exponent end cell row cell 2 to the power of open parentheses 2 straight x squared minus 2 straight x minus 4 close parentheses plus open parentheses straight x squared minus 5 straight x plus 4 close parentheses end exponent end cell less than cell 2 to the power of negative 4 end exponent end cell row cell 2 to the power of 3 x squared minus 7 x end exponent end cell less than cell 2 to the power of negative 4 end exponent end cell end table


Pertidaksamaan ini mempunyai bentuk a to the power of f left parenthesis x right parenthesis end exponent less than a to the power of m dengan a greater than 1, diperoleh a equals 2f left parenthesis x right parenthesis equals 3 x squared minus 7 x dan m equals negative 4. Penyelesaiannya adalah:


table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x squared minus 7 x end cell less than cell negative 4 end cell row cell 3 x squared minus 7 x plus 4 end cell less than 0 row cell left parenthesis 3 x minus 4 right parenthesis left parenthesis x minus 1 right parenthesis end cell less than 0 end table


Pembuat nol:


table attributes columnalign right center left columnspacing 0px end attributes row cell 3 x minus 4 end cell equals 0 row x equals cell 4 over 3 end cell end table


Atau:


table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 1 end cell equals 0 row x equals 1 end table


Penyelesaian:



Jadi, himpunan penyelesaiannya HP equals open curly brackets x vertical line 1 less than x less than 4 over 3 close curly brackets.

0

Roboguru

Nilai x yang memenuhi pertidaksamaan 125x​×3255x​&lt;625x−2​ adalah ....

Pembahasan Soal:

Ingat kembali bentuk persamaan eksponen berikut:

a to the power of f open parentheses x close parentheses end exponent less than a to the power of g open parentheses x close parentheses end exponent space rightwards arrow space f open parentheses x close parentheses less than g open parentheses x close parentheses space open square brackets untuk space a greater than 1 close square brackets 

Diketahui pertidaksamaan square root of 125 to the power of x end root cross times cube root of 25 to the power of 5 x end exponent end root less than square root of 625 to the power of x minus 2 end exponent end root, maka:

table attributes columnalign right center left columnspacing 0px end attributes row cell square root of 125 to the power of x end root cross times cube root of 25 to the power of 5 x end exponent end root end cell less than cell square root of 625 to the power of x minus 2 end exponent end root end cell row cell square root of open parentheses 5 cubed close parentheses to the power of x end root cross times cube root of open parentheses 5 squared close parentheses to the power of 5 x end exponent end root end cell less than cell square root of open parentheses 5 to the power of 4 close parentheses to the power of x minus 2 end exponent end root end cell row cell square root of 5 to the power of 3 x end exponent end root cross times cube root of 5 to the power of 10 x end exponent end root end cell less than cell square root of 5 to the power of 4 x minus 8 end exponent end root end cell row cell 5 to the power of fraction numerator 3 x over denominator 2 end fraction end exponent cross times 5 to the power of fraction numerator 10 x over denominator 3 end fraction end exponent end cell less than cell open parentheses 5 to the power of 4 x minus 8 end exponent close parentheses to the power of 1 half end exponent end cell row cell 5 to the power of fraction numerator 3 x over denominator 2 end fraction plus fraction numerator 10 x over denominator 3 end fraction end exponent end cell less than cell 5 to the power of 2 x minus 4 end exponent end cell row cell 5 to the power of fraction numerator 9 x over denominator 6 end fraction plus fraction numerator 20 x over denominator 6 end fraction end exponent end cell less than cell 5 to the power of 2 x minus 4 end exponent end cell row cell 5 to the power of fraction numerator 29 x over denominator 6 end fraction end exponent end cell less than cell 5 to the power of 2 x minus 4 end exponent end cell row cell fraction numerator 29 x over denominator 6 end fraction end cell less than cell 2 x minus 4 end cell row cell 29 x end cell less than cell 6 open parentheses 2 x minus 4 close parentheses end cell row cell 29 x end cell less than cell 12 x minus 24 end cell row cell 29 x minus 12 x end cell less than cell negative 24 end cell row cell 17 x end cell less than cell negative 24 end cell row x less than cell fraction numerator negative 24 over denominator 17 end fraction end cell end table 

Jadi, jawaban yang benar adalah A.

0

Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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