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Nilai x yang memenuhi pertidaksamaan 125x​×3255x​<625x−2​ adalah ....

Pertanyaan

Nilai x yang memenuhi 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 adalah ....

  1. x less than negative 24 over 17 

  2. x less than negative 22 over 17 

  3. x less than negative 19 over 17 

  4. x less than negative 25 over 11 

  5. x less than negative 23 over 11 

Pembahasan:

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.

Jawaban terverifikasi

Dijawab oleh:

W. Lestari

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 06 Oktober 2021

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Roboguru

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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