LOGARITHMS 21

Find x if Log_x0.000001 = -6

solution

Log_x0.000001 = -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 Log_x0.00000001 = -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  Log_aa^c = cLog_aa )

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

= 8 + 6

= 14

hence Log₂(256 x 64) = 14

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

Evaluate Log₂(512 x 128).