soal penilaian PAS
1). f (x) = k . 25x - 8
f (x) = k . 25 (2) - 8
20 = 50 k - 8
28 = 50k
k = 28/50
-3k = -3 . 28/50
= -84/50
2). y - 3/2-3 = x - 1/ 0-1
y - 3 / -1 = x - 1 / -1
-1 (y-3) = -1 (x-1)
-y + 3 = -x + 1
-y = -x - 2
y = x + 2
3). √8x²-4x+3 = 1/32^x-1
8 x²-4x+3/2 = 32^-x-1
3x²-12x+9/2 = -5x + 5
3x²-12x+19 = -10x + 10
3x² - 2x + 9 = 0
(3x -3/3) (3x + 1) = 0
(x-1) (3x+1) = 0
x = 1 atau x = -1/3
p+6q
1+6 (-1/3)
1 - 2 = -1
4). (2x-1)^8 = (-2 +x)^8
•> 2x - 1 = 0
x = 1/2
•> -2 + x = 0
x = 2
5). (2/3)^x = 6^1-x
log 2/3^x = log 6^1-x
x log2/3 = (1-x) log6
x log2/3 = log b - x log b
x log2/3 + x logb = log b
x (log 2/3 + log b) = log b
x = log b / log 2/3 + log 6
x = 1/1+log 1/9
6). (2x - 3)^x² - x = (2x -3)^x+4
•> f(x) = g(x)
x² - 2x = x + 4
x² - 3x - 4 = 0
(x - 4) (x + 1) = 0
x = 4 x = -1
•> 2x - 3 = 0
2x = 3
x = 3/2
•> 2x - 3 = -1
2x = 2
x = 1
•> 2x - 3 = 1
2x = 4
x = 2
7). (2x - 3)^x+1 = 1
(2x - 3)^x+1 = (2x - 3)^0
•> x + 1 = 0
x = 1
•> 2x - 3 = 0
x = 3/2
•> 2x - 3 = 1
x = 2
•> 2x - 3 = -1
x = 1
=> x1+x2+x3 = -1 + 3/2 + 2 + 1 + 3/2
= 2+3/2
= 5/2
8). 2^2x-6 × 2^+1 + 32
(2^x)² - 6 × 2^x × 2 + 32 = 0
p² - 12p + 32 = 0
(p - 8) (p - 4) = 0
p = 8 p = 4
•> p-8 → 2^x = 8
2^x = 2³
x = 3
•> p = 4 → 2^x = 4
2^x = 2²
x = 2
9). 3^2x+1 × 28 × 3^x + 9 = 0
(3^x)² × 3 - 28 × 3^x - 9 = 0
p² - 28p + 9 = 0
(3p - 27/3) (p-1)
(p = 9) (p = 1/3)
•> p = 9 → 3^x = 9
3^x = 3³
x = 3
•> p = 1/3 → 3^x = 1/3
3^x = 3^-1
x = -1
=> 3 X1 - X2
3 (3) - (-1) = 7
10). (5^x)² × 5 - 26 × 5^x+5 = 0
5p² - 26p + 5 = 0
(5p - 25/5) (5p - 1) = 0
(p - 5) (5p - 1) = 0
p = 5 p = 1/5
•> p = 5 → 5^x = 5¹
x = 1
•> p = 1/2 → 5^x = 5^-1
x = -1
=> X1 + X2 = 1 + (-1) = 0
11). 5^x²-2x-4 > 5^3x+2
x² - 2x - 4 > 3x + 2
x² - 5x - 6 > 0
(x - 6) (x+1) > 0
x = 6 x = -1
HP : {x < -1 atau x > 6}
12). (1/2)^2x-5 < (1/4)^½x + 1
(1/2)^2x-5 < (1/2)^2 (1/2 x + 1)
2x - 5 < x + 2
x < 7
13. Xo = 1.000.000 jiwa → tahun 2000
→ tahun 2001 = 1.040.000jiwa
→ tahun 2002 = 1.081.600 jiwa
→ tahun 2003 = 1.124.864 jiwa
14). Xo = 0,5 kg → pukul 08.00
→ pukul 09.00 = 0,49 kg
→ pukul 10.00 = 0,4802 kg
15). 5^x+2 < 4^x
log5^x+2 < log4^x
(x+2) log5 < x (log4)
xlog5 + 2log5 < xlog4
xlog5 - xlog4 < -2 log5
x (log5 - log4) < -2log5
HP = {x < -2log5 /log5-log4}
16). (x-4)^4x < (x-4)^1+3x
4x < 1+3x
x < 1
HP : {x < 1 atau x < 4}
x-4 < 0
x < 4
17). 2^x³-x < 1
2^x³ - x < 2^0
x³ - x < 0
x (x - 1) (x + 1) < 0
x = 0, x = 1, x = -1
HP : {0 < x < 1 atau x < -1}
18). 5^2x+1 > 5^x+4
(5^x)² × 5¹ > 5^x + 4
5 p² > p + 4
5p² - p - 4 > 0
(p - 1) (5p + 4) > 0
p = 1 p = -4/5
HP : {x < -4/5 atau x > 1}
19). 2^x × 2¹ × 2^-4 ≤ 0 / 1-2^x
p - 2p - 4 / 1-2x ≤ 0
(p-2) (p+1) / p (1-p) ≤ 0
p = 2 p = -1
HP : { -1 < x < 2}
20). (4^x)² × 4¹ > 4^x + 3
4p² - p - 3 > 0
(4p - 4/4) (4p +3) > 0
(p - 1) (4p + 3) > 0
p = 1 p = -3/4
HP : {-3/4 < x < 1}
21). 3^x-2y = 3^-4
x - 2y = -4
x - y = 4
__________________ -
-y = 8
y = -8
x-8 = 4
x = 12
x+y = 12 + (-8) = 4
22). (2a^5 - b^-5)/3 × 2a^9 × b^-1) ^-1
(2¹a^5 b^-5/2^5 a^9 b^-1)
(2^-4 a^-4 b^-4)
(2^⁴ a^⁴ b^⁴)
(2 ab)⁴
23). 9^3x-4 = 81^½x-5
(3²)^3x-4 = (3^-4)^2x-5
6x - 8 = -8x + 20
14x = 28
x = 2
24). 4¹+² × 3^4x+1 < 432
4 × 4^2x × 3^4x × 3 < 432
4^2x × 3^4x < 36
16^x × 18^x < 36
1.296^x < 36
36^x < 36
x < 1
25). (1/3)^x+2 < (1/3)^x
x + 2 < x
0 < 2 (TIDAK MEMENUHI)
(1/3)^x+2 < (1/3)^-x
x +2 < -x
2x < -2
x < -1
26). x = 0
27). Dari kelima fungsi yang diberikan pada opsi, hanya opsi E yang menunjukkan fungsi logaritma dengan 0 < a < 1
b = -2
c = 9
nilai minimum
y = ((-2)² -4(1)(9)) / (-4(1))
y = -32/-4
y = 8
log2ˣ + log5^z= log10¹⁰ + log3^y
log 2ˣ .5^z = log10¹⁰. 3^y
2ˣ . 5^z = 10¹⁰. 3^y
2ˣ . 5^z . 3^0= 2^10. 5^10. 3^y
x = 10
y = 0
z = 10
maka :
2x + 8y - 3z = 2(10) + 8(0) - 3(10)
= 20 + 0 - 30
= -10
30). ²logx² + ³logy⁻³ = 4
2²logx - 3³logy = 4misal ²logx=p, ³logy=q
maka, 2p-3q=4.... (1)
²logx + ³logy⁴ =13
⇒ ²logx + 4³logy=13
⇒ p+4q=13...(2)
subtitusikan pers.1 &2
2p - 3q =4
2p + 8q = 26
diperoleh
p = 5 ⇒ ²logx = 5
q = 2 ⇒ ³logy = 2
⁴logx -
31). ᵃlog b = n → b = aⁿ
²log (4ˣ + 6) = 3 + x4ˣ + 6 = 2³⁺ˣˣ₁
4ˣ + 6 = 2³. 2ˣ
(2ˣ)² - 8 (2ˣ) + 6 = 0
misal 2ˣ= a
a² - 8a + 6 = 0,
a1. a2 = 6
2ˣ₁. 2ˣ₂ = 6
2⁽ˣ₁⁺ˣ₂) = 6
x₁ + x₂ = ²log 6
xlog (4x + 12) = xlog x^2
berarti
4x + 12 = x^2
x^2 - 4x - 12 = 0
x = 6 atau -2
x²-5x+8-2 = 0
x²-5x + 6 = 0
(x-3) (x-2) = 0
x= 3 atau x = 2
6,9 - 3,2 = 3,7
Misal ²log x = a
a² - 3a - 10 = 0
(a - 5)(a + 2) = 0
a - 5 = 0 atau a + 2 = 0
a = 5 a = -2
²log x = a
²log x = 5
x = 2⁵
= 32
²log x = a
²log x = -2
x = 2⁻²
=
x₁.x₂ = 32 .
= 8
38). misalkan y = log x
maka bentuk persamaan menjadiy² - 4y + 3 = 0
(y - 1) (y - 3) = 0
y = 1 atau y = 3
y = 1
log x = 1
x = 10
y = 3
log x = 3
x = 10³
x = 1000
HP : {10 , 1000}
3x + 5 < 35
3x < 35 - 5
3x < 30
x < 30/3
x < 10
40). ²log (5x - 16) < 6
²log (5x - 16) < 2^6
5x - 6 < 64
5x < 70
x < 14
41). 4log (2x2 + 24) > 4log (x2 + 10x)
Syarat nilai pada logaritma.
2x2 + 24 > 0 (definit positif). Jadi, berlaku untuk setiap x . . . (1)
x2 + 10x > 0, maka x < -10 atau x > 0 . . . . (2)
Perbandingan nilai pada logaritma
(2x2 + 24) > (x2 + 10x)
2x2 - x2 - 10x + 24 > 0
x2 - 10x + 24 > 0
(x – 4)(x – 6) >
x < 4 atau x > 6 ....(3)
Jadi, dari (1), (2), dan (3) diperoleh penyelesaian x < -10 atau x > 6.
42) 43) 44). tidak paham :)
45). 2x - ⁵log (x² + 5x) > 2x - ⁵log (4x + 12)
0 > 2x - 2x + ⁵log (x² + 5x) - ⁵log (4x + 12)⁵log ((x² + 5x)/(4x + 12)) < 0
(x² + 5x)/(4x + 12) < 5⁰
(x² + 5x)/(4x + 12) - 1 < 0
(x² + 5x - 4x - 12)/(4x + 12) < 0
(x² + x - 12) / (4 (x + 3)) < 0
(x + 4) (x - 3) (x + 3) < 0
x + 4 = 0
x = -4
x + 3 = 0
x = -3
x - 3 = 0
x = 3
x < -4 atau -3 < x < 3
syarat
4x + 12 > 0
4x > -12
x > -3
x² + 5x > 0
x (x + 5) > 0
x < -5 atau x > 0
HP = { x | 0 < x < 3, x ∈ bilangan real }
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