Indices and Logarithms, Short Questions (Question 1 - 8)


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,   lo g 16 [ lo g 2 ( 5x 4 ) ]=lo g 9 3

Solution:
lo g 16 [ lo g 2 ( 5x 4 ) ]=lo g 9 3 lo g 16 [ lo g 2 ( 5x 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 lo g 2 ( 5x 4 )= 16 1 4 lo g 2 ( 5x 4 )=2 5x 4= 2 2 5x=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



Question 5
Solve the equation, log 9 ( x2 )= log 3 2

Solution:
log 9 ( x2 )= log 3 2 log a b= log c b log c a log 3 ( x2 ) log 3 9 = log 3 2 log 3 ( x2 ) 2 = log 3 2 log 3 ( x2 )=2 log 3 2 log 3 ( x2 )= log 3 2 2 x2=4 x=6



Question 6
Solve the equation, log 9 ( 2x+12 )= log 3 ( x+2 )

Solution:
log 9 ( 2x+12 )= log 3 ( x+2 ) log 3 ( 2x+12 ) log 3 9 = log 3 ( x+2 ) log 3 ( 2x+12 )=2 log 3 ( x+2 ) log 3 ( 2x+12 )= log 3 ( x+2 ) 2 2x+12= x 2 +4x+4 x 2 +2x8=0 ( x+4 )( x2 )=0 x=4 (not accepted) x=2



Question 7
Solve the equation, log 4 x= 3 2 log 2 3

Solution:
log 4 x= 3 2 log 2 3 log 2 x log 2 4 = 3 2 log 2 3 log 2 x 2 = 3 2 log 2 3 log 2 x=2× 3 2 log 2 3 log 2 x=3 log 2 3 log 2 x= log 2 3 3 x=27



Question 8
Solve the equation, 2 log 5 2 = log 2 ( 2x )

Solution:
2 log 5 2 = log 2 ( 2x ) 2= log 5 2. log 2 ( 2x ) 2= 1 log 2 5 . log 2 ( 2x ) 2 log 2 5= log 2 ( 2x ) log 2 5 2 = log 2 ( 2x ) 25=2x x=23