[tex] \\ [/tex]
[tex] \tt a. \: 3 \: log_{2}(2p) [/tex]
[tex] \tt b. log_{2}(2p) [/tex]
[tex] \tt c. \frac{3}{ \: log_{2}(2p) } [/tex]
[tex] \tt d. \frac{1}{ \: log_{2}(2p) }[/tex]
[tex] \tt e. \frac{3}{ \: log_{2}(p) }[/tex]
[tex] \\ [/tex]
○ Wajib sertai langkah pengerjaan,
○ Dilarang menyali (copas) *+warn
Jawab:
C. = 3÷(log_{2} (2p)) (p ≠ ½) ✔️
Penjelasan dengan langkah-langkah:
Ini cara Cambridge,
bilangan pokok ditaruh di kanan bawah log
[[ alumni SMA Cambridge ]]
Note: _{x} = bilangan pokok x
Diketahui ∵ log_{z} (y) = x -> zˣ = y ∴
--> log_{p} (a) = 2 --> p² = a
--> log_{q] (8p) = 2 --> q² = 8p
Ditanya nilai log_{2p} (pq² ÷ a)
∵ q² = 8p --> p = q²/8
p² = a --> (q²/8)² = a --> a = q⁴/64 ∴
Nilai log_{2p} (p·q² ÷ a)
= log_{2p} ((q²/8)·q² ÷ (q⁴/64))
= log_{2p} ((q²q²/8) ÷ (q⁴/64))
= log_{2p} ((q²⁺²/8) ÷ (q⁴/64))
= log_{2p} ((q⁴/8) ÷ (q⁴/64))
∵ pembagian, nilai pembilang penyebut
harus dibalikkin ∴
= log_{2p} ((q⁴/8) × (64/q⁴))
maka
= log_{2p} (q⁴/q⁴ × (64/8))
= log_{2p} (64/8)
= log_{2p} (8)
= log_{2p} (2³)
∵ log_{a] (bⁿ) = n·log_{a} (b) ∴
= 3log_{2p} (2)
∵ log_{a} (b) = 1/(log_{b} (a)) ∴
= 3÷(log_{2} (2p)) (opsi c) ✔️
log_{2}(2p) ≠ 0
2⁰ = 1, 2p = 1, p = ½
maka, p ≠ ½
[[ KLF ]]
[answer.2.content]