LOGARITHMS 3

Evaluate log₂1024⁸

Solution

= log₂1024⁸

= log₂(2¹⁰)⁸

= log₂2⁸⁰

= 80 x log₂2    [since log_aa = 1]

= 80 x 1

= 80

Hence log₂1024⁸ = 80

TRY THIS…………………..

Evaluate log₂2048¹⁰

DISTANCE 1

LOGARITHMS 2

If Log_ax = ¹/₂  Log_a16+ ¹/₃ Log_a125 find x.

Solution

Log_ax = ¹/₂  Log_a16+ ¹/₃ Log_a125

Log_ax = Log_a16^1/2  +  Log_a125^1/3   

Log_ax = Log_a 4 +  Log_a5

Log_ax = Log_a (4 x 5)

Log_ax = Log_a 20

x = 20

TRY THIS………………

If Log_ax = ¹/₃  Log_a216+ ¹/₄ Log_a625 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

S_n = 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

=  m^8x2

=  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  Log_aa^c = cLog_aa )

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

= 2 + 8

= 10

hence Log₂(4 x 256) = 10

TRY THIS……………..

Evaluate Log₂(4 x 16).