LOGARITHMS 16

If Log(30x-90/5+x) = 1. Find x.

Solution

Log(30x-90/5+x) = 1.

Log₁₀(30x-90/5+x) = 1.

 30x-90 = 10¹. After changing into exponential form

   5+x

30x-90 = 10    since [10¹=10]

   5+x

30x-90 = 10(5+x)    after cross multiply

  

30x-90 = 50+10x

30x-10x = 50+90 [collecting like terms]

20x = 140

x=7  [after dividing by 20 both sides]

Hence x=7.

TRY THIS……………

Log(30x-120/3+x) = 1.

FACTORIZE 3

Factorize 25 – (x-3)²

Solution

= 25 – (x-3)²

= 5² – (x-3)²   Let a=5 and b=(x-3)

= [5 - (x-3)][5 + (x-3)]     Since a²-b²=(a-b)(a+b)

= [5 – x+3][5 + x-3]  

= [5+3 – x][5 – 3 + x]   

 

= [8 – x][2 + x] 

   

= (8 – x)(2 + x) answer   

TRY THIS…………….. 

Factorise 81 – (x-2)²

PROBABILITY 1

The probability that Azam FC will beat Yanga in the national league is ⁹/16. Find the probability that it won’t be able to do so.

Solution

Let P(E)=  ⁹/16         P(E’) = ?

P(E)   + P(E’) = 1

⁹/16 + P(E’) = 1

P(E’) = 1 -  ⁹/16

P(E’) = ¹⁶/16 -  ⁹/16

P(E’) =  16-9

               16

P (E’) =  7

              11

Hence answer is ⁷/11.

TRY THIS………………………

The probability that Mtibwa FC will beat Mbeya City FC in the national league is ¹³/40. Find the probability that it won’t be able to do so.

POLYGONS 2

Find the exterior angle of a regular polygon with 6 sides.

Solution

n = 6

e =   360

         n

e =   360

         6

e =   360 ⁶⁰

        6₁

e = 60⁰

Hence the interior angle is 60⁰

TRY THIS………………………………………..

Find the exterior angle of a regular polygon with 20 sides.

FACTORIZE 2

Factorize 25m²n² - c²

Solution

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

=25m²n² - c²

= 5²m²n² - c²

= (5mn)² - (c)²

= (5mn - c)( 5mn + c)

Hence 25m²n² - c² = (5mn - c)( 5mn + c) answer

TRY THIS…………….

Factorize 64a²d² - e²

EXPAND 1

Expand (7x + 3)(x - 2)

Solution

= (7x + 3)(x - 2)

= x(7x + 3)-2(7x + 3)

 = 7x² + 3x - 14x - 6

 = 7x² - 11x – 6 

TRY THIS………………

Expand (8x + 5)(x - 3)

FUNCTIONS 6

Find remainder when P(x) = x⁵ + 3x³ – x +6  is divided by Q(x)=x+2;

Solution

Let x+2 = 0

x=0-2

x = -2 

Now we substitute x=-2 in P(x).

P(x) = x⁵ + 3x³ – x +6 

P(-2) = (-2)⁵ + (-2)³ – (-2) +6 

P(-2) = -32 + (-8) – (-2) +6 

P(-2) = -32 + (-8) + 2 + 6  

P(-2) = -40 + 8 

P(-2) = -32

So remainder is -32

TRY THIS……………

Find remainder when P(x) = x⁵ - 2x³ – x + 9  is divided by Q(x)=x-3;

EVALUATE 1

Evaluate  90 + 6 – 5 – 7 – 9 + 15.

Solution

= 90 + 6 – 5 – 7 – 9 + 15

= 90 + 6 + 15 – 5 – 7 – 9

= 90 + 6 + 15 – ( 5 + 7 + 9)

= 111 – 21

= 90

TRY THIS………

Evaluate  75 + 11 – 17 – 9 – 13 + 15

EXPONENTIALS 3

Simplify 300

               64^-2/3

Solution

=  300

      64^-2/3

=  300 x 1

             64^-2/3

=  300 x 64^2/3        [Since 1/a^-n = aⁿ]

=  300 x (64^1/3)²        [Since 64^1/3 = cube root of 64 = 4]

=  300 x (4)²        

=  300 x 16

= 4800        

         

Hence   300       = 4800

              64^-2/3

TRY THIS…………………………

simplify 11

             27^-2/3

INEQUALITIES 1

Find x if 5x – 29 ≤ x + 15 ≤ 6x.

Solution

5x – 29 ≤ x + 15 and  x + 15 ≤ 6x

5x – x ≤ 29+ 15 and  15 ≤ 6x - x

4x ≤ 44 (divide by 4 both sides) and  15 ≤ 5x (divide by 5 both sides)

x ≤ 11 and  3 ≤ x

x ≤ 11 and  x ≥ 3 answer

TRY THIS……..

Find x if 6x – 55 ≤ x + 15 ≤ 2x. 

LOGARITHMS 15

Evaluate Log100 +  log 0.001 - log 0.0001+ log 0.00001

Solution

= Log100 + log 0.001 -  log 0.0001+ log 0.00001

= Log10² +  log 10^-3 – log 10^-4 + log 10^-5

= 2Log10 + (-3 log 10) – (-4log 10) + (-5log 10)

= (2x1) + (-3x1) - (-8x1) + (-5 x 1)

= 2 + (-3) – (-4) -5

= 2 - 3 + 4 - 5

= 2 + 4 - 3 - 5

= 6 - 8

= 2  

Hence Log100 + log 0.001-  log 0.0001+ log 0.00001 = 2

TRY THIS……………………… 

Evaluate Log100 +  log 100,000 - log 0.0001+ log 0.00001

WORD PROBLEMS 3

The difference between two numbers is 21 and their sum is 683. Find the two numbers.

Solution

Let the two numbers be x and y.

x – y = 21 ----------- (i)

x +y = 683 ------------(ii)

---------------------------------------------

From (i)

 x – y = 21

x = 21+ y -------------- (iii)

-------------------------------------------

Substitute (iii) in (ii) above.

x +y = 683. ------------(ii)

(21+ y) +y = 683

21 + 2y = 683

2y = 683 - 21

2y = 662

¹2y = 662³³¹

₁2      2₁

y = 331

from (iii)

x = 21+ y

x = 21+ 331

x = 352

Hence the two numbers are 331 and 352.

TRY THIS…………………………………   

The difference between two numbers is 33 and their sum is 167. Find the two numbers.

WORD PROBLEMS 2

When 7 is subtracted from twice a certain number, the result is greater than 93. Find the number.

Solution

Let the number be w.

2w – 7 > 93

2w > 93 + 7

2w > 100

2w > 100

2        2

w > 50 answer

TRY THIS…………………………… 

When 30 is subtracted from thrice a certain number, the result is greater than 76. Find the number.

RATIOS 1

Decrease 87000sh by 3:10.

Solution

= 3 x 87000sh

   10

= 3 x 87000⁸⁷⁰⁰

   10₁

= 3 x 8700

= 26,100Tshs answer

TRY THIS………………………………………… 

Decrease 28,000sh by 7:20.

SIMPLIFY 4

Simplify 20 + 6(100n ÷ ⁿ/2) - 80

Solution

=20 + 6(100n ÷ ⁿ/2) - 80

=20 + 6(100n x ²/n) – 80

=20 + 6(100n¹ x ²/n₁) – 80  cancelling n

=20 + 6(100 x  2) – 80 

=20 + 6(200) – 80 

=20 + 1200 – 80 

=1220 – 80 

=1140

=1140answer

TRY THIS……..

Simplify 70 + 4(10n ÷ ⁿ/5) - 35

WORD PROBLEMS 1

Ana, Joy and Fred shared 2400 sweets such that Ana got 50 sweets less than Joy while Fred got three as much as what Ana got. Find those given to Fred.

Solution

The starting point is Joy. Let the sweets given to Joy be x. See the table below:

ANA	JOY	FRED
x-50	x	3(x-50)=3x-150

  

Then;

ANA + JOY + FRED = 2400

x-50 + x + 3x-150 =2400

x + x + 3x - 150 - 50 = 2400

5x - 200 = 2400

5x = 2400 + 200

5x = 2600

5x = 2600

5       5

x = 520

The sweets given to Fred

= 3x – 150

= 3(520) – 150

= 1560 – 150

=1410

Hence Fred got 1410 sweets.

TRY THIS……………………………………… 

Kemi, Roy and Isaya shared 246 sweets such that Kemi got 10 sweets less than Roy while Isaya got four as much as what Kemi got. Find those given to Isaya.   [132 answer]

FUNCTIONS 5

If F(x) = 4x + 13; Find F^-1(x).

Solution

HINT: F^-1(x) means inverse.

PROCEDURE:

Make x the subject and then interchange x and y variables.

Let y=F(x)

So,  y= 4x + 13

y – 13 = 4x

y – 13  = 4x

   4          4

y – 13   = x

   4

x  =   y – 13     after rearranging

           4

F^-1(x) = x – 13    after interchanging x and y variables.

                4

Hence, F^-1(x) = x – 13     

                              4

TRY THIS………………   

If F(x) = 7x + 20; Find F^-1(x).

SIMPLIFY 3

Simplify  8w(5w – 7 + w) + 12w².

solution

= 8w(5w – 7 + w) + 12w².

= (8w x 5w) – (8w x 7) + (8w x w) + 12w².

= [40w² – 56w + 8w² ]+ 12w².

= 40w² + 8w² + 12w² – 56w

= (48w² + 12w²)  – 56w

= 60w² – 56w

= 60w² – 56w answer

TRY THIS………..

Simplify 3a(7a+ 30 - a) +29a².

FACTORIZE 1

Factorize 81 - 225m²

Solution

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

81 - 225m² = 9² - 15²m²       

                = 9² – (15m)²       

                = (9 - 15m)(9 + 15m)

Hence 81 - 225m² = (9 - 15m)(9 + 15m) answer

TRY THIS…………….

Factorize 49 – 121y²