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Materi Matematika

materi

PERSAMAAN LOGARITMA

alog f(x) = alog g(x) f(x) = g(x)

alog f(x) = b f(x) =ab

f(x)log a = b (f(x))b = a

Dengan syarat x yang didapat dari persamaan tersebut harus terdefinisi. (Bilangan pokok > 0 1 dan numerus > 0 )

Contoh:

Tentukan nilai x yang memenuhi persamaan berikut !

xlog 1/100 = -1/8
x-1/8 = 10-2
(x -1/8) -8 = (10-2)-8
x = 10 16

xlog 81 - 2 xlog 27 + xlog 9 + 1/2 xlog 729 = 6
xlog 34 - 2 xlog33 + xlog + 1/2 xlog 36 = 6
4 xlog3 - 6 xlog3 + 2 xlog3 + 3 xlog 3 = 6
3 xlog 3 = 6
xlog 3 = 2
x = 3
x = 3 (x>0)

xlog (x+12) - 3 xlog4 + 1 = 0
xlog(x+12) - xlog 4 = -1
xlog ((x+12)/4) = -1
(x+12)/4 = 1/x
x + 12x - 64 = 0
(x + 16)(x - 4) = 0
x = -16 (TM) ; x = 4

logx - 2 logx - 3 = 0

misal : log x = p

p - 2p - 3 = 0
(p-3)(p+1) = 0

p1 = 3
log x = 3
x1 = 2 = 8

p2 = -1
log x = -1
x2 = 2-1 = 1/2

Bilangan pokok a > 0 1

Tanda pertidaksamaan tetap/berubah tergantung nilai bilangan pokoknya

a > 1

0 < a < 1

a log f(x) > b f(x) > ab
a log f(x) < b
f(x) < ab

(tanda tetap)

a log f(x) > b f(x) < ab
a log f(x) < b
f(x) > ab

(tanda berubah)

syarat f(x) > 0


Contoh:

Tentukan batas-batas nilai x yang memenuhi persamaan

log(x - 2x) < 3
a = 2 (a>1)
Hilangkan log Tanda tetap


- 2 < x < 0 atau 2 < x < 4

x - 2x < 2
x - 2x -8 < 0
(x-4)(x+2) < 0
-2 < x < 4

syarat : x - 2 > 0
x(x-2) > 0
x < 0 atau x > 2

1/2log (x - 3) < 0
a = 1/2 (0 < a < 1)
Hilangkan log Tanda berubah


x < - 2 atau x > 2

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