5.2.3 Indices, PT3 Practice


Question 14:
Solve:  3 2n+1 = 3 n ×9 ( 9 1 2 ) 3

Solution:
3 2n+1 = 3 n ×9 ( 9 1 2 ) 3 3 2n+1 = 3 n × 3 2 3 3 3 2n+1 = 3 n+2( 3 ) 2n+1=n+5        n=4


Question 15:
Given that  k 3 = 9 3 2 × 64 1 2 , find the value of k.

Solution:
k 3 = 9 3 2 × 64 1 2    = ( 3 2 ) 3 2 × ( 2 6 ) 1 2    = 3 3 × 2 3    = ( 2 3 ) 3 k= 2 3


Question 16:
Given 9 x+2 ÷ 3 4 = 3 x+1 , calculate the value of x.

Solution:
9 x+2 ÷ 3 4 = 3 x+1 ( 3 2 ) x+2 ÷ 3 4 = 3 x+1     3 2x+4 ÷ 3 4 = 3 x+1   2x+44=x+1              2x=x+1                x=1


Question 17:
Simplify:  ( 2 x 5 y 2 z 1 6 ) 3 ÷ 1 x 2 z

Solution:
( 2 x 5 y 2 z 1 6 ) 3 ÷ 1 x 2 z = 8 x 15 y 6 z 1 2 × x 2 z = 8 x 15 y 6 z 1 2 × ( x 2 z ) 1 2 = 8 x 15 y 6 z 1 2 ×x z 1 2 =8 x 15+1 y 6 = 8 x 16 y 6

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