Pertanyaan

Jika x 1 d an x 2 memenuhi persamaan , maka x 1 x 2 = ....

Jika $x_{1} d an x_{2}$ memenuhi persamaan $open parentheses 2 log invisible function application x minus 1 close parentheses. fraction numerator 1 over denominator log presubscript space presuperscript x invisible function application 10 end fraction equals log invisible function application 10$ , maka $x_{1} x_{2 = ....}$

Ikuti Tryout SNBT & Menangkan E-Wallet 100rb

Habis dalam

Klaim

N. Rahayu

Master Teacher

Mahasiswa/Alumni Universitas Negeri Jakarta

Jawaban terverifikasi

Pembahasan

$open parentheses 2 log invisible function application x minus 1 close parentheses. fraction numerator 1 over denominator log presubscript space presuperscript x invisible function application 10 end fraction equals log invisible function application 10 open parentheses 2 log invisible function application x minus 1 close parentheses. open parentheses log invisible function application x close parentheses equals log invisible function application 10 M i s a l k a n space log invisible function application x equals a open parentheses 2 a minus 1 close parentheses open parentheses a close parentheses equals 1 2 a squared minus a equals 1 2 a squared minus a minus 1 equals 0 open parentheses 2 a plus 1 close parentheses open parentheses a minus 1 close parentheses equals 0 D i d a p a t space a equals negative 1 over 1 space a t a u space a equals 1 log invisible function application x equals negative 1 half left right double arrow x subscript 1 equals 10 to the power of negative 1 half end exponent log invisible function application x equals 1 left right double arrow x subscript 2 equals 10 to the power of 1 D i d a p a t space x subscript 1 x subscript 2 equals 10 to the power of negative 1 half plus 1 end exponent equals 10 to the power of 1 half end exponent equals square root of 10$