LOGARITMA

Logaritma merupakan invers dari bentuk pangkat. jika a^n=b, maka dalam logaritma dinyatakan menjadi n = a'log b, dengan a disebut basis. misalnya 2^3 = 8 dalam logaritma 2'log 8 = 3.

catatan:

untuk basis 10, dalam logaritma tidak dituliskan, misalnya log 3 artinya 10'log 3

Sifat-Sifat Logaritma:

1. a'log a = 1

2. a'log 1 = 0

3. a'log (a^n) = n

4. a'log (b^n) = n.(a'log b)

5. a'log (b.c) = (a'log b) + (a'log c)

6. a'log (b/c) = (a'log b) - (a'log c)

7. (a^m)'log (b^n) = (n/m).(a'log b)

8. a'log b = 1/(b'log a)

9. (a'log b).(b'log c).(c'log d) = a'log d

10. a'log b = (c'log b)/(c'log a)

11. a^(a'log b) = b

contoh soal:

*hitunglah nilai dari logaritma-logaritma berikut,

a} 2'log 6

b} 8'log 16

c} 36'log 128, jika 9'log 18 = m - 1 

d} 35'log 98 , jika 2'log 5 = a dan 7'log 2 = b

penyelesaian:

a} 2'log 6

= 2'log (2.3)

= (2'log 2) + (2'log 3)

= 1 + (2'log 3)

b} 8'log 16

= (2^3)'log (2^4)

= (4/3).(2'log 2)

= 4/3

c} 9'log 18 = m - 1

9'log (2.9) = m - 1

1 + (9'log 2) = m - 1

9'log 2 = m - 2

1/(2'log 9) = 1/(m-2)

36'log 128

= 1/(128'log 36)

= 1/((2^7)'log (4.9))

= 1/[((2^7)'log 4)+((2^7)'log 9)]

= 1/[(1/7)(2+(2'log 9))]

= 1/[(1/7)(2+(1/(m-2)))]

= 1/[(1/7)((2m-3)/(m-2))]

= 1/[(2m-3)/7(m-2)]

= [7(m-2)] / [2m-3]

d} 35'log 98

= [log 98]/[log 35]

= [2(log 7)+(log 2)]/[(log 5)+(log 7)]

= [2(2'log 7)+(2'log 2)]/[(2'log 5)+(2'log 7)]

= [2(1/b)+1]/[a+(1/b)]

= [(2+b)/b]/[(ab+1)/b]

= (2 + b) / (ab + 1)