Saturday 28 February 2015

LOGARITHMS 3


Evaluate log210248

Solution

= log210248

= log2(210)8

= log2280

= 80 x log22    [since logaa = 1]

= 80 x 1

= 80

Hence log210248 = 80

TRY THIS…………………..


Evaluate log2204810

DISTANCE 1





LOGARITHMS 2


If Logax = 1/2  Loga16+ 1/3 Loga125 find x.

Solution

Logax = 1/2  Loga16+ 1/3 Loga125

Logax = Loga161/2  +  Loga1251/3   

Logax = Loga 4 +  Loga5

Logax = Loga (4 x 5)

Logax = Loga 20

x = 20

TRY THIS………………


If Logax = 1/3  Loga216+ 1/4 Loga625 find x.

SIMULTANEOUS EQUATIONS 1


Solve the following equations by substitution method.
    2x - 3y = 6
    x + 5y = 16

Solution

2x - 3y = 6 ----------- (i)
x + 5y = 16 ---------- (ii)

From equation (ii)

x + 5y = 16
x = 16- 5y ----------(iii)

substitute equation (iii) in (i) above,

2x - 3y = 6

2(16- 5y) - 3y =6
32 – 10y – 3y = 6
32 – 13y = 6
– 13y = 6 - 32
– 13y =  - 26

13y =  - 26
  -13        -13

y = 2

From equation (iii)

x = 16- 5y
x = 16- 5(2)
x = 16- 10
 x = 6

Hence x=6 and y=2.

TRY THIS……………

Solve the following equations by substitution method.
4x - 2y = 6

x + y = 9

GEOMETRICAL PROGRESSION 1


Find the sum of the 1st six terms of the geometrical progression 2+6+18+54+…….

Solution

G1 = 2, r=3, n=6

Sn = G1(rn-1)/ r-1
               
S7 = 2(36-1) /3-1
             
S7 = 12(36-1)
        21

S7 = (36-1)

S7 = 729 - 1

S7 = 728

Hence the sum of the 1st seven terms is 728.

TRY THIS……………………………


Find the sum of the 1st seven terms of the geometrical progression 2+8+32+…….

EXPONENTS 1


Simplify (m8)2

solution

= (m8)2

=  m8x2

=  m16

TRY THIS........


Simplify  (h10)3

BODMAS 1


Evaluate  20 + 10 – 8 – 7 – 9 + 14.

Solution

= 20 + 10 – 8 – 7 – 9 + 14

= 20 + 10 + 14 – 8 – 7 – 9

= 20 + 10 + 14 – (8 + 7 + 9)

= 44 – 24

= 20

TRY THIS………..


Evaluate  20 + 10 – 7 – 8 – 6 + 24

RADICALS 2




ARITHMETICS 1


Evaluate 96 x 856 + 9144 x 96.

Solution

= 96 x 856 + 9144 x 96

= 96 x (856 + 9144)

= 96 x 10,000

= 960,000

Hence 96 x 856 + 9144 x 96 = 960,000

TRY THIS…………


Evaluate 35 x 745 + 9255 x 35

RADICALS 1



LOGARITHMS 1


Evaluate Log2(4 x 256).

Solution

= Log2(4 x 256)

= Log24 +  Log2256          (applying the product rule)

= Log222 +  Log228             ( 4= 22 and 256=28 )

= 2Log22 +  8Log22       ( remember  Logaac = cLogaa )

= (2 x 1) +  (8 x 1)          ( remember  Logaa = 1 )

= 2 + 8

= 10

hence Log2(4 x 256) = 10

TRY THIS……………..


Evaluate Log2(4 x 16).