EXPONENTIALS 1

If a² = 11; find the value of a⁶.

solution

a⁶ = (a²)³.

a⁶ = (11)³.   [ since a² = 11]

a⁶ = 11 x 11 x 11

a⁶ = 1331   answer

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

If c² = 7; find the value of c⁶.

LOGARITHMS 1

Simplify Log₂8192 – Log₅625

Solution

= Log₂8192 – Log₅625

= Log₂2¹³ – Log₅5⁴ 

= 13Log₂2 – 4Log₅5   because Log_aaⁿ =  nLog_aa

= (13 x 1) – (4 x 1)      because Log_aa =1

= 13 – 4

= 9

Hence Log₂8192 – Log₅625 = 6 

TRY THIS…………..

NECTA 1995 QN 24b

Simplify Log₂32 – Log₃9.

FACTORIZE 1

Factorize 121x²- 16y²

solution

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

121x²- 16y² = 11²x² - 4²y²

                = (11x)² - (4y)²

                = (11x - 4y)( 11x + 4y)

Hence 121x²- 16y² = (11x - 4y)(11x + 4y)

TRY THIS…………….

Factorize 49m²- 100n²

EXPONENTIALS 1

If y² = 3, find y¹⁰- y⁴

Solution

= y¹⁰- y⁴

= (y²)⁵ - (y²)²  

= (3)⁵ - (3)²     remember y² = 3

= 243  - 9   

= 234

Hence y¹⁰- y⁴ = 234.

TRY THIS……………….

If  y⁷ = 10, find y²¹- y¹⁴

MID-POINT 1

Find a if the midpoint of a line from (a, 6) to (4, 10) is (7, 7)

Solution

x1=a, x2=4, y1=6, y2=10.

(7, 7)= (x1 + x2, y1 + y2)

                 2             2

(7, 7)= (a + 4,  6 + 10)

                2          2

(7, 7)= (a + 4, 16)

                2      2

Equating values of x;

7= a + 4

        2      

2 x 7= (a + 4)  x 2 ¹

              2 ₁     

14 = a + 4

14-4 = a

a = 10.

Hence a=10

TRY THIS……………………..

Find a if the midpoint of a line from (a, 12) to (7, 10) is (3, 11)