Assalamualaikum, hari ini saya akan membahas tentang soal-soal PAS beserta cara dan jawaban nyaa....
Informasi : Tanda (^) = pangkat.
Tanda (.) = kali
Tanda (/) = per
(2,20)= f(2) = 20
K . 2^5 (2)-8 = 20
K . 4 = 20
K = 5
-3k = -3(5)= -15
2. Y = a.b^2 + a simbol
( 1,3 ) , (0,2) , (2,5)
(1,3) = 3 = a.b+c
3 = 1.b+ 1
3 = b + 1
b = 2
(0,2) = 2 = a.b^2 + c
2 = a + 1
a = 1
Jadi, a = 1
b = 2
c = 1
y = 1. 2^x + c
y = 2^x + 1
3. = 2^3 ( x^2 - 4x + 3)/2 = 2^-5(x-1)
= 3(x^2-4x+3) = -10(×-1)
= 3x^2 - 2x - 1 = 0
= (3x + 1) (x-1)= 0
X = -1/3 x = 1
P > a
P = 1
a = -1/3
P + 6a = 1 + 6(-1/3)
= - 1
4.( 2x-1)^8 = (-2+x)^8 ➡️ karena pangkat genap, maka persamaan dimutlakan (2x-1) = (-2+x)
2x - 1 = -2 + x
2x - x = -1
Atau,
2x - 1 = 2 - x
2x + x= 2+1
3x = 3
X = 1
jadi, Himpunan penyelesaian nya { - 1, 1 }
5. ( 2/3)^x = 6^1-x
➡️log 2^x/3 = log 6^1-x
X. Log 2/3 = 1-x .log 6
Log 2/3 / log6 = 1-x/x
6 log 2/3 = 1/x - 1
6 log 2/3 + 1 = 1/x
^6 log 2/3 + ^6log 6 = 1/x
^6 log 4 = 1/x
1/^6 log 4 = x
X = ^4 log 6
6.(2x - 3 )^x^2-2x = (2x-3)^x14
X^2-2x = x + 4
X^2 - 3x - 4 = 0
(X-4) (x+1)
X = 4 X = -1 HP { -1,4 }
7. (2x-3)^x+1
X + 1 = 0
X1 = -1
2x - 3 = 1
X2 = 2
2x - 3 = -1
X3 = 1
(X1 + X2 + X3 ) = (-1,2,1)
= -1 + 2 + 1
= 2
8. 2x - 6.2^x+1 + 32 = 0
➡️(2^x)^2 - 12 (2^x) + 32 = 0
(2^x - 6) (2^x - 4) = 0
2^x = 6 ➡️ X1 = 3
2^x = 4 ➡️ X2 = 2
Jadi, 2X1 + X2
2 (3) + 2
= 6 + 2
= 8
9. 3^2x. 3^1 - 28.3^x + 9 = 0
(3^x)^2.3 - 28 . 3^x + 9 = 0
Misal = 3^x = p
3p^2 - 25p + 9 = 0
( 3p - 1 ) (p - 9)
P = 1/3 p = 9
➡️3^x . 1/3
3^x = 3^-1
X = -1 ➡️ X2
➡️3x = 9
3^x = 3^2
X = 2 ➡️ X1
Jadi, X1 > X2
2 > -1
= 3(2) - -(1)
= 6 + 1
= 7
10. 5^2x+1 - 26. 5^x + 5 = 0
➡️ 5^2x . 5 - 26 .5^x + 5 = 0
Misal 5^x = a
5a^2 - 26a + 5 = 0
(5a - 1) (a - 5)
a = 1/5 a = 5
➡️ 5^x = 1/5 5^x = 5
5^x = 5^-1 5^x - 5^1
-1 + 1 = 0
11. 5^x2-2x-4 > 5^3x+2
➡️ x^2 - 2x - 4 > 3x + 2
X^2 - 5x - 6 > 0
( x - 6 ) ( x + 1 )
X = 6 X = -1
12. (1/2)^2x-5 < (1/4)^1/2x + 1
(2^-1)^2x +5 < (2^-2)^1/2x+1
2^-2+5 < 2^-x-2
-2^x + 5 < -x - 2
-x < -7
X > 7
13. Soal pertumbuhan
Dik = y0 = 1.000.000
r = 4% = 0,04
x = 2003 - 2000 = 3
Dit = y ..?
y = y0 ( 1 + r ) ^x
= 1.000.000 ( 1 + 0,04)^3
= 1.124.864
14. Soal peluruhan
Dik = y0 = 0,5 kg
r = 2 %
X = 10.00 - 8 00 = 2
Dit = y...?
y = y0 (1 - r )^x
= 0,5 (0,98)^2
= 0,5 (0,9604)
= 0,4802
15.5^x+2 < 4^x
log 5^x+2 < log 4^x
(X+2) log 5 < (x) log 4
X+2/x < ^5 log 4
X/X + 2/x < ^5 log 4
1 + 2 /x < ^5 log 4
2/x < ^5 log 4
2/ x < ^5 log 4 - 1
2/x < ^5 log 4 - ^5 log 5
2/x < ^5 log(4/5)
2 < x.^5 log (4/5)
X > 2/^5 l9g (4/5)
X > (4/5) log (25)^-1
X > 5/4 log 1/25
16. (X - 4 ) ^ 4x < (x - 4)^ 1+ 3x
Langkah 1
➡️ f(x) = g(x)
4x < 1+3x
X < 1 ✅
Langkah 2
➡️h(x) = g(x)
X-4 < 1
X < 1 + 4
X < 5 ✅
Langkah 3
➡️h(x) = -1
X - 4 < 1
X < - 1+4
X < 3✅
⬇️
Uji f(x) dan g (x)
f(x) = 4x g(x) = 1 + 3x
= 4(3) = 1 + 3(3)
= (-1)^12 = (-1)^10
Langkah 4
h(x) = 0 ( dengan syarat x > 0 )
X - 4 < 0
X < 4✅
⬇️
Uji f(x) dan g(x)
f(x) = 4x g(x) = 1 + 3x
=4(4) = 1 + 3(4)
= 16 = 13
Jadi, Himpunan penyelesaian nya {1,5,3,4}
17. 2^3-x < 1
2^x3-x < 2^0
X^3 - X < 0
X(x^2-1) < 0
X ( x + 1) (x-1) < 0
Jadi, Himpunan penyelesaian nya { x1 < -1 atau 0 < x < 1}
18. 5^2x+1 > 5^x+4
5^2x .5^1-5^x-4 > 0
(5^x)^2.5 - (5^x) -4 > 0
5a^2 - a - a > 0
(5a + 4) (a -1 )
5a = -4 a=1
a = -4/5 5^x = 1
5^x = -4/5 5^x = 5^0
x = 0
Jadi, Himpunan penyelesaian nya = x > 0
19. 2^x - 2^1 - 1/ 1-2^x < 0
Misal x = 2
2^2. 2^1-2/1-2^2
= 4 - 1/2 - 1/ 1-4
= -5/6
2^x-2^1-x = 0
X = 1
1-2^x
X = 0
Jadi,Himpunan penyelesaian nya { x<0 atau x>1 }
20. 4^2x+1 > 4^x+3
4^2x.4 > 4^x + 3
(4^x)^2.4 > 4^x+3
Misal 4^x = P
P^2.4 > 4 + 3
4P^2 > P + 3
4P^2 - P - 3 > 0
(X - 4) (X+3)
X = 4 X = -3
Jadi,Himpunan penyelesaian nya { X|X < -3
atau X > 4}
21. 3^x-24 . 3^-4
2^x - 4 = 16 ➡️ 2^x-4 = 2^4
( X - 2y = -4) - ( X - y = 4)
= -y = -8 - y = 8
X - y = 4
X - 3 = 4
X = 12
X + y = 12 +8
= 20
22. ( 2a^a b^-9/32a^3b^-1) = (1/16a^4-3
b^9+1)^-1
= (1/6a^-1b^-8)
= 16ab^8
23. 9^3x - 4 = 1/81^2x-5
9^3x-4 = 81^-2 + 5
9^3x-4 = (9^2) - 2x + 5
3x-4 = -4x + 10
3x+4x = 10 + 4
7x = 14
X = 2
24. 4^1+2x . 3^4x+1 < 432
4^1+2x . 3^4x+1 < 16.27
4^1+2x . 3^4x+1 < 2^4 . 3^3
(2^2)^1+2.3^4x+1 < 2^4 . 3^3
2^2+4x . 3^4x+1 < 2^4 . 3^3
2 + 4x < 4
4x < 4-2
4x < 2
X < 1/2
25. (1/9)^x+2 < (1/3)^x
(1/3^2)^x+2 < (1/3)^x
(3^-2)^x+2 < 3^-x
3^-2x-4 < 3^-x
2x - 4 < -x
-2x + x < 4
-x < 4
X > 4
26. F(x) = 3.^2 log(3x)
ubah F(x) menjadi 0, sehingga kita peroleh
3.^2log(3x) = 0
^2log(3x) = 0
3x = 2
3x = 1
jadi nilai f bernilai 0 adalah x = 1/3
27. Hanya e. F(x) = 1/2 log x+4 termasuk fungsi turunan. Karena 0< a< 1, yaitu 1/2 = 0,5
,jadi f(x) = 0,5log x+4
28. F(x) =^2log(x^2- 2x + 9)
nilai basis (a=2) lebih besar dari 1, sehingga
f(x) minimum tercapai numerus
g(x) = x^2-2x+9 karena g(x) adalah fungsi kuadrat maka nilai minimum g(x)
X = -b/2a
maka, X = (-2)/2(1) = 1
subtitusi X = 1 pada f(x)
F(1) = ^2log (x^2 - 2x + 9 )
= ^2log (1)^2 - 2(1)+9)
= ^2log 8
= 3
jadi,nilai minumum nya adalah 3.
29. 2 - ylog 3+2 log 5= 10 maka 2x+8y -3z=
X(log 2) - y(log 3) + 2(log5) = 10
2^x - 3^y + 5^z = 10^10
30. ^2log^2x + ^3log^y3 = 4
➡️ 2^2log x - 3^3 log y = 4
maka 2log x = p , 3log y = 4
maka 2p - 3p = 4 ....(1)
^2 log x + ^3log y^q = 13
➡️ ^2log x + 4^3 log y = 13
p + 4q = 13 ....(2)
subtitusi 1 & 2
2p - 3p = 4
p + 4q = 13
Diperoleh P = 5 ➡️ 2log x = 5
q = 2 ➡️ 3 log x = 2
➡️ 4log x - y log 9
4log x - 1/3^2log y
1/2 log x - 2/3log y
5/2 - 2/2 = 3/2
31. alog b=n ➡️ b = an
2log (4^x+6) = 3+x
4x+6 = 2^3+xx
4x+6 = 2^3.2^x
(2^x)^2 - 8(2^x)+6 = 0
misal 2^x = a
a^2 - 8a + 6 = 0 , akar a1 dan a2
a1. a2 = 6
2^x 1 . 2^x 2 = 6
2^(x + x)
1 2 = 6
x1 + x2 = ^2 log 6
32. X > 0 dan X tidak sama dengan 1
syarat numerus
x log (4x + 12 ) = 2
x^2 = 4x + 12
x^2 - 4x - 12 = 0
( x - 6) ( x + 2) = 0
X = 6 atau X = -2
nilai x yang memenuhi kedua syarat diatas hanya untuk X = 6
33. 2log (x^2 - 5x + 8 )= ^2log 2
X^2 - 5x + 8 = 2
X^2 -5x + 8-2 = 0
X^2 - 5x + 6 = 0
(X - 3) ( X - 2) = 0
X = 3 atau X = 2
34. 2log 48/3 + 5log 50/2
2 log 16 + 5 log 25
4 + 2 = 6
35. 2 log 125/9 = 2 log 5^3 - 2 log 3^2
3 (2 log 5) - 2 ( 2log 3)
3(2,3) - 2(1,6) = 6,9 - 3,2 = 3,7
36). soal tidak lengkap
37). ²Log² x - 3.²log x - 10 = 0
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 menjadi
y² - 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}
39). 5 log 3x+5 < 5 log35
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 }πΊπππππ π
πππ ππππ,wassalamualikum πΊπππππ ππππππππ ππ!
π¦πππ π£π’π€π πππ π€ππ¬ππ₯ππ‘ππ§..
