Thursday 30 April 2015

LOGARITHMS 21




Find x if Logx0.000001 = -6

solution

Logx0.000001 = -6

0.000001  = x-6 .  [0.000001  = 10-6  ]

10-6 = x-6 .

106 = x6 .  { equal powers cancel out}

x = 10

Hence x = 10

TRY THIS..................

Find x if Logx0.00000001 = -8

FACTORIZATION 3




Factorize x2 - 6x - 16 = 0

solution

x2 - 6x - 16 = 0    we split -6x to get ( -8x + 2x)

( -8x + 2x) = -5x    and  ( -8x x 2x) = -16x2

x2 - 6x - 16 = 0   

(x2 -8x) + (2x - 16) = 0

x(x - 8) + 2(x - 8) = 0

 (x - 8)(x +2) = 0

x - 8= 0 or x + 2 = 0

x = 8 or x = -2

TRY THIS.....................

Factorize x2 - 9x - 22 = 0    

FUNCTIONS 7




If f(x) = x4 + kx2 + 3x + 8 has a remainder of 22 when divided by x+2; find k.

solution

f(x) = x4 + kx2 + 3x + 8

x + 2 = 0
x = -2

f(x) = (-2)4 + k(-2)2 + 3(-2) + 8 = 22

16 + 4k + (-6) + 8 = 22

16 + 4k + 2 = 22

4k + 18 = 22

4k = 22 - 18

4k = 4

4k = 4
4      4

k = 1

hence k=1

TRY THIS......................

If f(x) = x4 - kx2 + 3x - 8 has a remainder of 40 when divided by x+3; find k.

DIFERRENCE OF 2 SQUARES - 7




Factorize 1-m2

Solution

We use difference of two squares a2 – b2 = (a - b)(a + b)

1-m2 = 12-m2       

        = (1 - m)(1 + m)

Hence 1 - m2 = (1 - m)(1 + m)


TRY THIS…………….


Factorize 16 - y2

LOGARITHMS 20





Evaluate Log2(256 x 32).

Solution

= Log2(256 x 64)

= Log2256 +  Log264          (applying the product rule)

= Log228 +  Log226             ( 256= 28 and 32=25 )

= 8Log22 +  6Log22       ( remember  Logaac = cLogaa )

= (8 x 1) +  (6 x 1)          ( remember  Logaa = 1 )

= 8 + 6

= 14


hence Log2(256 x 64) = 14

TRY THIS...............

Evaluate Log2(512 x 128).