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soal #1
jika log 2 = p, tentukan nilai
log 156,25 dalam p!
soal #2
tentukan nilai maksimum & minimum dari fungsi :
f(x) = √7 cos (1/7π) – 2√7
##Quiz #23 ##
Quiz #23
soal1
log 2 = p
log 10 = log 2 + log 5 = 1
log 5 = 1 – p
log 156,25
= log 156 1/4
= log 625/4
= log 5⁴ – log 2²
= 4 log 5 – 2 log 2
= 4(1 – p) – 2p
= 4 – 6p
soal2
nilai maksimum cos ax = 1
nilai minimum cos ax = -1
soal sy edit
f(x) = √7 cos (1/7 x) – 2√7
Nilai maks = √7 . 1 – 2√7 = – √7
Nilai min = √7 . (-1) – 2√7 = – 3√7
1.
LOGARITMA
log (156,25) = log(¼(156,25 × 4))
= log(625) -log(4)
= log(10.000/16) -log(2²)
= log(10.000) -log(16) -2 log(2)
= log(10⁴) -log(2⁴) -2p
= 4 -4 log(2) -2p
= 4 -4p -2p
= 4 -6p
2.
GRAFIK FUNGSI TRIGONOMETRI
f(x) = √7 cos(⅐x) -2√7
f(x) = (cos(⅐x) -2)√7
nilai maksimum terjadi jika cos(⅐x) = 1
= (1 -2)√7
= -√7 => nilai maksimum
nilai minimum terjadi jika cos(⅐x) = -1
= (-1 -2)√7