LOGARITHMS G1

 

Evaluate Log₂(512 ÷ 8).

 

Solution

 

= Log₂(512 ÷ 8)

 

= Log₂512 -  Log₂8          (applying the product rule)

 

= Log₂2⁹ -  Log₂2³             ( 512= 2⁹ and 8=2³ )

 

= 9Log₂2 -  3Log₂2       ( remember  Log_aa^c = cLog_aa )

 

= (9 x 1) -  (3 x 1)          ( remember  Log_aa = 1 )

 

= 9 - 3

 

= 6 answer

 

 

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

 

Evaluate Log₂(2048 - 32).

ALGEBRA G1

 

Given that one of the roots of the equation 3x² + m(x+1) + 5 = 0 is 2, find m.

Solution

Substitute x=2, in the above equation.

3(2)² + m(2+1) + 5 = 0

(3x4) + (m x 3) + 5 = 0

12 + 3m + 5 = 0

3m + 17 = 0

3m = -17

3m = -17

3          3

 

m=^-17/3

Hence m=^-17/8

TRY THIS…………………………..

Given that one of the roots of the equation 4x² + m(x+1) + 10 = 0 is 8, find m.

GEOM. PROGRESSION G1

 The 1st term of a geometric progression is 7 and the 5th term is 112. Find i)the common ratio. ii) 10^th term

 

Solution

 

G₁=7, G₅=112, n=5

 

i) G_n = G₁r^n-1;

 

G₅ = G₁r^5-1;

 

G₅ = G₁r⁴;

 

112 = 7 x r⁴;

 

112 = 7r⁴;

 7       7

 

16 = r⁴;  Finding the fourth root of 16,

 

r = 2

 

 

 

Hence the common ratio is 2.

 

ii)  10^th term

 

G_n = G₁r^n-1;

 

G₁₀ = G₁r^10-1;

 

G₁₀ = G₁r⁹;

 

G₁₀ = 7 x 2⁹;

 

G₁₀ = 7 x 512;

 

G₁₀ = 3584

 

Hence the eighth term is 3584

 

TRY THIS………………………….. 

 

The 1st term of a geometric progression is 2 and the 6th term is 486. Find i) the common ratio. ii) 11^th term.

 

EXPONENTIALS H2

 

FUNCTIONS H1

 Find a linear function f(x) with gradient -20 which is such that f(5)=16.

 

Solution.

 

m=-20, points = [5,16] and [x, f(x)]

 

m = y₂-y₁/x₂-x₁

 

-20 = [f(x) – 16]/x-5

 

f(x)-16=-20(x-5)  [after cross multiplying]

 

f(x)-16=-20x+100

 

f(x)=-20x+100+16

 

f(x)=-20x+116

 

A  linear function is f(x) = -20x + 116.

 

 

TRY THIS……….

 

 

Give out a linear function f(x) with gradient -17 and f(2)=28

ARITHMETIC PROGRESSION H1

 

The 1st term of an A.P. is 78 and the common difference is 45. Find the 10th term.

 

Solution

 

A₁= 78, d = 45

 

A_n = A₁ + (n-1)d

 

A₁₀ = A₁ + (10-1)d

 

A₁₀ = A₁ + 9d

 

A₁₀ = 78 + (9x45) [ after substituting A₁= 78, d = 45 as given above]

 

∴ A₁₀ = 78 + 405

 

A₁₀ = 483

 

Hence the 10th term is 483.

 

  

TRY THIS……………

 

 

The 1st term of an A.P. is 140 and the common difference is 23. Find the 18th term.

ALGEBRA H1

 

Simplify the expression r² - t²/ (r + t)

 

Solution

 

from difference of two squares p² - q² = (p - q)(p + q)

 

= r² - t²/ (r + t)

 

= (r + t)(r - t)

      r + t

 

= (r + t)(r - t)

      r + t

 

=  r - t    answer

      

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

 

 

Simplify the expression 

 

w² - y²/ (w + y)

EXPONENTIALS H1

 

If y⁵ = 3, find y¹⁰ – y²⁰.

 

Solution

 

= y¹⁰ – y²⁰

 

= (y⁵)² - (y⁵)⁴

 

= (3)²  - (3)⁴          remember y² = 3

 

= 9  -  (3x3x3x3)

 

= 9 – 81

 

= -72

 

Hence y¹⁰ – y²⁰ = -72.

 

TRY THIS……………….

 

If  y⁹ = 10, find y¹⁸ – y²⁷.

ARITHMETIC H1

 Evaluate 22 x 237 + 463 x 16.

 

Solution

 

= 222 x 237 + 463 x 222

 

= 222 x (237 + 463)     factoring out the common number

 

= 222 x 700   

 

= 155400      [after multiplying 222 and 7 and adding two zeros on the answer]

 

Hence 222 x 237 + 463 x 222 = 155400

 

 

TRY THIS………….

 

 

Evaluate 124 x 516 + 484 x 124.

COMPOUND INTEREST H1

 

Salva invested a certain amount of money in a bank which gives an interest rate of 10% compounded annually. How much did she invest at the start if she got 300,000 sh at the end of 4 years?

 

solution

 

n=4, T=1, R=10%, A₄=300,000, P=?

 

A_n = P(1 + ^RT/100)ⁿ

 

A₄ = P(1 + ^(10x1)/100)⁴

 

A₄ = P(1 + ¹⁰/100)⁴

 

A₄ = P(¹⁰⁰/100 + ¹⁰/100)⁴

 

A₄ = P(¹¹⁰/100)⁴  But A₄=8000,

 

300,000 = P(1.1)⁴  

 

300,000 = 1.4641P

 

300,000  =  1.4641P

1.4641        1.4641

 

300,000  =  P    [use math tables to divide if you like]

1.4641

 

P = 204,904.04 [to 2 d.p]

 

 

Hence at the start she invested Tsh 204,904.04

 

TRY THIS……………………..

 

Sylvia invested a certain amount of money in a bank which gives an interest rate of 10% compounded annually. How much did she invest at the start if she got 150,000 sh at the end of 5 years?