5.6 Indices and Logarithms, SPM Practice (Long Questions)


5.6 Indices and Logarithms, SPM Practice (Long Questions)
Question 1:
(a)  Find the value of
        i.      2 log2 12 + 3 log25 – log2 15 – log2 150.
     ii.      log832
(b)  Shows that 5n  + 5n + 1 + 5n + 2  can be divided by 31 for all the values of n which are positive integer.

Solution:
(a)(i)
2 log2 12 + 3 log2 5 – log215 – log2 150
= log2 122 + log2 53– log2 15 – log2 150
= log 2 12 2 × 5 3 15 × 150
= log2 8
= log2 23
= 3

(a)(ii)
log 8 32 = log 2 32 log 2 8             = log 2 2 5 log 2 2 3 = 5 3

(b)
5n  + 5n + 1 + 5n + 2
= 5n  + (5 × 5n ) + (52 × 5n )
= 5n  (1 + 5 + 52)
= 31 × 5n  

Therefore, 5n  + 5n + 1 + 5n + 2 can be divided by 31 for all the values of n which are positive integer.


Question 2:
(a)  Given log10 x = 3 and log10y = –2. Shows that 2xy – 10000y2 = 19.
(b)  Solve the equation log3 x = log9(x + 6).

Solution:
(a)
log10x = 3      → (x = 103)
log10y = –2    → (y = 10-2)
2xy – 10000y2 = 19
Left hand side:
2xy – 10000y2
= 2 × 103 × 10-2 – 10000 (10-2)2
= 20 – 10000 (10-4)
= 20 – 1
= 19
= right hand side

(b)
log 3 x = log 9 ( x + 6 ) log 3 x = log 3 ( x + 6 ) log 3 9 log 3 x = log 3 ( x + 6 ) log 3 3 2 log 3 x = log 3 ( x + 6 ) 2
2log3 x= log3 (x + 6)
log3 x2= log3 (x + 6)
x2= x + 6
x2x – 6 = 0
(x + 2) (x – 3) = 0
x = – 2 atau 3.
log3 (– 2) not accepted (logarithm of a negative number is undefined)
Jadi, x = 3.

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