Short Questions (Question 1 – 4)


Question 1
Solve the equation, log3 [log2(2x – 1)] = 2

Solution:
log3 [log2 (2x – 1)] = 2 ← (if log a N = x, N = ax)
log2 (2x – 1) = 32
log2 (2x – 1) = 9
2x – 1 = 29
x = 256.5




Question 2
Solve the equation,   l o g 16 [ l o g 2 ( 5 x   4 ) ] = l o g 9 3

Solution:
l o g 16 [ l o g 2 ( 5 x   4 ) ] = l o g 9 3 l o g 16 [ l o g 2 ( 5 x   4 ) ] = 1 4 log 9 3 = log 9 3 1 2 = 1 2 log 9 3 = 1 2 ( 1 log 3 9 ) = 1 2 ( 1 2 ) = 1 4 l o g 2 ( 5 x   4 ) = 16 1 4 l o g 2 ( 5 x   4 ) = 2 5 x   4 = 2 2 5 x = 8 x = 8 5



Question 3
Solve the equation, 5 log 4 x = 125

Solution:
5 log 4 x = 125 log 5 5 log 4 x = log 5 125 put log for both side ( log 4 x ) ( log 5 5 ) = 3 ( log 4 x ) ( 1 ) = 3 x = 4 3 = 64




Question 4
Solve the equation, 5 log 5 ( x + 1 ) = 9

Solution:
5 log 5 ( x + 1 ) = 9 log 5 5 log 5 ( x + 1 ) = log 5 9 log 5 ( x + 1 ) . log 5 5 = log 5 9 log 5 ( x + 1 ) = log 5 9 x + 1 = 9 x = 8


5.3 Equations Involving Indices


METHOD:
  1. Comparison of indices or base
    1. If  the base are the same , when a x = a y , then x = y
    2. If  the index are the same , when a x = b x , then a = b

  2. Using common logarithm (If base and index are NOT the same)
a x = b lg a x = lg b x = lg b lg a


Example 1 (Index Equation - Equal base)
Solve each of the following.
(a) 16 x = 8
(b) 9 x .3 x 1 = 81
(c) 5 n + 1 = 1 125 n 1









Example 2 (Solving index equation simultaneously)
Solve the following simultaneous equations.

2 x .4 2 y = 8

3 x 9 y = 1 27