Pertanyaan

Tentukanlah nilai x supaya: 4 × 4 2 × 4 3 × ... × 4 14 2 × 2 2 × 2 3 × ... × 2 x = 1

Tentukanlah nilai $x$ supaya:

$\frac{2 \times 2 ^{2} \times 2 ^{3} \times ... \times 2 ^{x}}{4 \times 4 ^{2} \times 4 ^{3} \times ... \times 4 ^{14}} = 1$

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nilai yang tepat adalah .

nilai

yang tepat adalah

Pembahasan

Diketahui: Ditanya: Nilai Perlu diingat bahwa: Perhatikan perhitungan berikut: Dengan menggunakan deret aritmatika yaitu , didapat: Karena nilai tidak mungkin bernilai negatif, maka nilai adalah . Jadi, nilai yang tepat adalah .

Diketahui:

$fraction numerator 2 cross times 2 squared cross times 2 cubed cross times... cross times 2 to the power of x over denominator 4 cross times 4 squared cross times 4 cubed cross times... cross times 4 to the power of 14 end fraction equals 1$

Ditanya:

Nilai

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 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

Perhatikan perhitungan berikut:

$table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 2 cross times 2 squared cross times 2 cubed cross times... cross times 2 to the power of x over denominator 4 cross times 4 squared cross times 4 cubed cross times... cross times 4 to the power of 14 end fraction end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over 4 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over 4 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over open parentheses 2 squared close parentheses to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent over 2 to the power of 2 plus 4 plus 6 plus 8 plus 10 plus... plus 28 end exponent end cell equals 1 row cell 2 to the power of 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end exponent end cell equals cell 2 to the power of 2 plus 4 plus 6 plus 8 plus 10 plus... plus 28 end exponent end cell row cell 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end cell equals cell 2 plus 4 plus 6 plus 8 plus 10 plus... plus 28 end cell row cell 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end cell equals cell 2 open parentheses 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 close parentheses end cell end table$

Dengan menggunakan deret aritmatika yaitu S subscript n equals n over 2 open parentheses a plus U subscript n close parentheses , didapat:

x plus 21 equals 0 space atau space x minus 20 equals 0 x equals negative 21 space atau space x equals 20

Karena nilai tidak mungkin bernilai negatif, maka nilai adalah .

Jadi, nilai yang tepat adalah .