Tentukanlah nilai supaya:

Pertanyaan

Tentukanlah nilai $x$ supaya:

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

Pembahasan Soal:

Diketahui:

Ditanya:

Nilai $x$

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:

$table attributes columnalign right center left columnspacing 0px end attributes row cell 1 plus 2 plus 3 plus 4 plus 5 plus... plus x end cell equals cell 2 open parentheses 1 plus 2 plus 3 plus 4 plus 5 plus... plus 14 close parentheses end cell row cell x over 2 open parentheses 1 plus x close parentheses end cell equals cell 2 times 14 over 2 open parentheses 1 plus 14 close parentheses end cell row cell x open parentheses x plus 1 close parentheses end cell equals cell 2 times 14 times 15 end cell row cell x squared plus x end cell equals 420 row cell x squared plus x minus 420 end cell equals 0 row cell open parentheses x plus 21 close parentheses open parentheses x minus 20 close parentheses end cell equals 0 end table$
$x plus 21 equals 0 space atau space x minus 20 equals 0 x equals negative 21 space atau space x equals 20$

Karena nilai $x$ tidak mungkin bernilai negatif, maka nilai $x$ adalah $20$ .

Jadi, nilai $x$ yang tepat adalah $20$ .

Pembahasan terverifikasi oleh Roboguru

Terakhir diupdate 18 Juli 2021

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Ruangguru

Download di Google Play

Download di Appstore

Download di App Gallery

Produk Ruangguru

Produk Lainnya

Bantuan & Panduan

Hubungi Kami

081578200000

[email protected]

02140008000

Ikuti Kami

Instagram

Facebook

Twitter

Youtube