Tentukan himpunan penyelesaian pertidaksamaan berikut. a. (41​)2x−3>322−x

Pertanyaan

Tentukan himpunan penyelesaian pertidaksamaan berikut.

a. open parentheses 1 fourth close parentheses to the power of 2 x minus 3 end exponent greater than 32 to the power of 2 minus x end exponent 

A. Allamah

Master Teacher

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himpunan penyelesaiannya adalah open curly brackets x space vertical line space x greater than 4 close curly brackets.

Pembahasan

Ingat beberapa sifat bilangan berpangkat berikut:

  1. a to the power of negative m end exponent equals 1 over a to the power of m
  2. 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

Dengan menggunakan kedua sifat di atas, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 1 fourth close parentheses to the power of 2 x minus 3 end exponent end cell greater than cell 32 to the power of 2 minus x end exponent end cell row cell open parentheses 1 over 2 squared close parentheses to the power of 2 x minus 3 end exponent end cell greater than cell open parentheses 2 to the power of 5 close parentheses to the power of 2 minus x end exponent end cell row cell open parentheses 2 to the power of negative 2 end exponent close parentheses to the power of 2 x minus 3 end exponent end cell greater than cell open parentheses 2 to the power of 5 close parentheses to the power of 2 minus x end exponent end cell row cell 2 to the power of negative 4 x plus 6 end exponent end cell greater than cell 2 to the power of 10 minus 5 x end exponent end cell end table

Kemudian, pada pertidaksamaan eksponen berlaku:

Jika a to the power of f open parentheses x close parentheses end exponent greater than a to the power of g open parentheses x close parentheses end exponent, maka:

  • f open parentheses x close parentheses greater than g open parentheses x close parentheses, untuk a > 1.
  • f open parentheses x close parentheses less than g open parentheses x close parentheses, untuk 0 < a < 1.

Berdasarkan teorema di atas, maka diperoleh:

table attributes columnalign right center left columnspacing 0px end attributes row cell 2 to the power of negative 4 x plus 6 end exponent end cell greater than cell 2 to the power of 10 minus 5 x end exponent end cell row cell negative 4 x plus 6 end cell greater than cell 10 minus 5 x end cell row cell negative 4 x plus 5 x end cell greater than cell 10 minus 6 end cell row x greater than 4 end table 


Dengan demikian, himpunan penyelesaiannya adalah open curly brackets x space vertical line space x greater than 4 close curly brackets.

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