Roboguru

Tentukan penyelesaian pertidaksamaan:

Pertanyaan

Tentukan penyelesaian pertidaksamaan:

open vertical bar 4 over 2 to the power of x minus 3 end exponent close vertical bar less or equal than 8

Pembahasan Soal:

Diketahui:

open vertical bar 4 over 2 to the power of x minus 3 end exponent close vertical bar less or equal than 8

Ditanya:

Penyelesaian pertidaksamaan open vertical bar 4 over 2 to the power of x minus 3 end exponent close vertical bar less or equal than 8

Perlu diingat bahwa:

a to the power of m times a to the power of n equals a to the power of m plus n end exponent space space 1 over a to the power of m equals a to the power of negative m end exponent Jika space a less than 1 space dan space a to the power of f left parenthesis x right parenthesis end exponent less than a to the power of g left parenthesis x right parenthesis end exponent space maka space f left parenthesis x right parenthesis less than g left parenthesis x right parenthesis 

Perhatikan perhitungan berikut:

open vertical bar 4 over 2 to the power of x minus 3 end exponent close vertical bar less or equal than 8 minus 8 less or equal than 4 over 2 to the power of x minus 3 end exponent less or equal than 8 fraction numerator negative 8 over denominator 8 end fraction less or equal than fraction numerator 4 over denominator 8 times 2 to the power of x minus 3 end exponent end fraction less or equal than 8 over 8 minus 1 less or equal than fraction numerator 1 over denominator 2 times 2 to the power of x minus 3 end exponent end fraction less or equal than 1 minus 1 less or equal than 1 over 2 to the power of x minus 3 plus 1 end exponent less or equal than 1 minus 1 less or equal than 1 over 2 to the power of x minus 2 end exponent less or equal than 1 minus 1 less or equal than 2 to the power of negative open parentheses x minus 2 close parentheses end exponent less or equal than 1 minus 1 less or equal than 2 to the power of negative x plus 2 end exponent less or equal than 1 minus 1 less or equal than 2 to the power of negative x end exponent times 2 squared less or equal than 1 minus 1 less or equal than 2 to the power of negative x end exponent times 4 less or equal than 1 fraction numerator negative 1 over denominator 4 end fraction less or equal than fraction numerator 2 to the power of negative x end exponent times 4 over denominator 4 end fraction less or equal than 1 fourth fraction numerator negative 1 over denominator 4 end fraction less or equal than 2 to the power of negative x end exponent less or equal than 1 fourth fraction numerator negative 1 over denominator 4 end fraction less or equal than 1 over 2 to the power of x less or equal than 1 fourth

Karena pangkat x tidak mungkin bernilai negatif, maka menjadi:

fraction numerator negative 1 over denominator 4 end fraction less or equal than 1 over 2 to the power of x less or equal than 1 fourth 0 less than 1 over 2 to the power of x less or equal than 1 fourth 1 over 2 to the power of x less or equal than 1 over 2 squared 2 to the power of negative x end exponent less or equal than 2 to the power of negative 2 end exponent minus x less or equal than negative 2 x greater or equal than 2 

Jadi, penyelesaian pertidaksamaan open vertical bar 4 over 2 to the power of x minus 3 end exponent close vertical bar less or equal than 8 adalah open curly brackets x space left enclose space x greater or equal than 2 comma space x element of straight R end enclose close curly brackets.

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

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