ROOT AND POWER

When a quantity m is raised to power in. It means that the quantity m should be multiple in the number of times of n 

Mⁿ = m₁ x m₂ x m₃...mₙ

2³ = 2 x2 x2 

From the above examples we can see that power gives the number of times the base is to be used as a factor. Power is also known to be an index or exponent.

Base -------X ʸ⁻⁻⁻⁻⁻⁻ power ( index)

8² mean 8 is raised to the power of 2 or 8 required 

z³ = means z cubed

2⁵  means fifth power of two

*Rule of  powers*

1. When a negative base is raised to an even power. it gives a positive result.

a.  (-2)² = (-2) x (-2)      =+4   

b. (-3)⁴ = (-3) x (-3) x (-3) x (-3) = +81

ii. when a negative base is raised to an odd number, it gives a negative result 

a.  (-2)³ = (-2) x (-2) x (-2) = -8

b. (-3)⁵ = (-3) x (-3) x (-3) x (-3)x (-3) = -243

iii. When a positive base is raiaed to either even or odd number, the result is positive.

a.  (-2)³ = (-2) x (-2) x (-2) = -8

b. (-3)⁵ = (-3) x (-3) x (-3) x (-3)x (-3) = -243

iii. When a positive base is raiaed to either even or odd number, the result is positive.

a. 2³ = 2 x 2 x 2 = 8

b.  3⁴ = 3 x 3 x3 x3 = 

     81

When any base ( except zero ) is raised to zero power , the result is positive one

a. x⁰ = 1

b. (-10)⁰ = 1

note 0⁰  = 1

When a base is raised to a negative power, the result is reciprocal of the base of the  base of the original number raised to the positive of that  power

a⁻ˣ = 1/aˣ

  *Worked examples*

Evaluate the following

I.   4² = 4 × 4 = 16 

II.  (-4)² = -4 × -4 = 16

III. 2² x (-3)³

  = -2 x -2 x(-3)(-3)(-3)

= -4 x (-27)

= 108

iv. (3)⁻² x (-2)3

2. Given that x=3, y= -1, and z = -2. Evaluate

x³ + y³ + z³

(3)³ + (-1)³ + (-2)³

27 + (-1) + (-8)

 = 18

ROOT

A root is obtained when a base is raised to a positive fraction power. The fractional power forms. Study the below question carefully

WORKED EXAMPLES

Find the value of y in each of the following 

Z^1/n  = ⁿ√(Z)

Where ( ^) means  ofpower of

1. 8^⅓ = ³√(8) 

= ³√(2 x 2 x 2) 

= 2

Example 2

What is the value of 

y⁴ = 81

add cube root to both side

⁴√(y⁴) = ⁴√(81)

Note: its only a fourth root that can cancel fourth power of y .We need to establish (y) alone

Hence,

y = ⁴√(81)

y = ⁴√(3 x 3 x3 x3)

y = ⁴√(3)⁴

y = 3

now you can see that its the fractional power of a number that forms its root and that the root of a number equals (=) to  one of its equal factors. 

*Rules of roots*

Given that y = ⁿ√(z)

a.   If n is a even number and y > 0, y will have two real solution, but if n is odd, y will have only two solution

Exampe 3

y² = 9 

add squared roots to both side

and so y = 

√(9) = + 3 or -3

4.   y³ = 64 

add cube root to both side

³√(y³) = ³√(64)

y = ³√(4 x 4 x4)

y = ³√(4³)

and so y 

y = 4

b. If n is an even number and y <0, z will have no real root, but if n is an odd number and y < 0, z will have only one z (negative) root

Example 5

a. y² = -4, and so y = √(-4) = No real root

b. y³ = - 8, 

and so x

 = ³√(8) =

- 2

Example 7

i √(7056)

 √(16 x 441)

√(16) x √(441)

4 x 21 = 84

ii. √(14400)

√(441 x 100)

= √(441) x √(100)

= 12 x 10 =120

iii. ³√(-2744)

³√(-8 x 343)

= ³√(-8) x ³√(343)

     -2 x  7 = -14

Example 8

if p = -7 and q = -2, find

a. (p-q)(p+q)

[(-7)-(-3)][(-7) + (-2)]

(-7 +3)(-7 - 2)

(-4)(-9)

=36

 

b.  p² - 2pq + q²

= (p -q)(p-q)

  [ (-7)-(-2)][(-7)-(-2)]

 

(-7 +2)(-7 +2)

= (-5)(-5) = 25

c. p + 2q - p²

(-7) + (2 x -2) - (-7)²

-7 + (-4) - (49)

-7 -4 -49

-60

SEE LINEAR INEQUALITIES

LINEAR INEQUALITY

Inequality had been part of life when two things differs in content, values privilege and quantity when it should be on the same ground State. The phenomenon called INEQUALITY had been set in place

Inequalities means ordering. The order or position of 2 in relation to 3 or 0, in the real number system, can be described as follows: 2 is less than 3 and 2 is greater than 0.

The symbol '<' is used to mean 'strictly less than' , and the symbol '>' means 'strictly greater than'. The symbol '≥' means 'greater than or equal to' while the symbol '≤' means 'less than or equal to'.

PROPERTIES OF INEQUALITIES

If a, b and c are real numbers then the following relation hold:

1. a < b⇔b - a > 0 , where the symbol '⇔' means the expression on the right is true if the previous statement holds.

2. One and only one of a < b, a = b , a > b is always true

3. If a > 0 and b > 0 , then a + b >0

4. If a >0 and b > 0 , then ab > 0

5. If a < b , there exists a 'c' such that a < c < b

6. If 'a' is given , then we can find some b and c such that b < a < c.

Example 1: If a, b and c ∈ R and a < b , where R is the set of real numbers, prove that

(i) a + c < b + c

(ii) a - c < b - c

(iii) ac < bc if c > 0

(iv) ac > bc if c < 0

Solution

(i) a < b ⇔ b - a > 0 by property 1 

But ( b + c ) - ( a + c ) = b - a for any c

( b + c ) - ( a + c ) > 0

b + c > a + c

An inequality is preserved if a fixed quantity is added to each side of it

(ii) Since a < b ⇒ b - a > 0, the symbol '⇒' is read as 'implies that;

     Then ( b - c ) - ( a - c ) = b - a

    ( b - c ) - ( a - c ) > 0

    b - c > a - c

An inequality is preserved if a fixed quantity is subtracted from each side of it.

(iii) Since a < b ==> b - a > 0

      Then for any c > 0, ( b - a )c > 0

      bc - ac > 0

     bc > ac

Multiplying each side of an inequality by a fixed positive quantity doesn't alter the inequality.

(iv) If a < b and c < 0 then

      b - a > 0 and ( -c ) > 0

     ( b - a )( -c ) > 0

     -bc + ac > 0

     ac > bc

Multiplying each side of an inequality by a fixed negative quantity reverses the inequality.

Example 2: Find the range of values of x for which each of the following will hold:

(i) ¾x - 2 < 4

(ii) ⅔( x - 1 ) > ½( 1 - 2x ) + ½

Solution

(i) ¾x - 2 < 4

     Multiplying both sides by 4, gives 3x - 8 < 16

     ⇔ 3x - 8 + 8 < 16 + 8

      ⇔ 3x < 24

           x < 8

(ii) ⅔( x - 1 ) > ½( 1 - 2x ) + ½

      Multiplying both sides by 6

     4( x -1 ) > 3( 1 - 2x ) + 3

    4x - 4 > 3 - 6x + 3

    10x > 10

     x > 1

Example 3: Solve the inequality 4x - 3 ≥ 2x + 3

Solution

4x - 3 ≥ 2x + 3

⇔ 4x - 2x - 3 + 3 ≥ 2x - 2x + 3 + 3

 ⇔ 2x ≥ 6 or x ≥ 3

STANDARD FORM

  A number is in its standard form if it's in the form (A ×10ⁿ),where A is a number between 1 and 10 and (n ) is a power of 10.

    Standard form is a way of writing very large or small numbers easily. 

 For example,the number 6000,can be written as 6 ×10³

Since 10³ =1000.

When very small numbers are written in standard form,their power or index are negative.

For example,0.1 can be written as

1×10⁻¹.

   

WORKED EXAMPLES

Write the following in Standard form.

1) 75,000,000 

    This is greater than 1. You rewrite it as :

  75,000,000.0

Now, move the decimal point 

7 places to the left to get a number between 1 and 10,which is 7.

    = 7.5×10,000,000

     = 7.5 × 10⁷

2) 0.000 0012

   

  solution:

0.000 0012

ᵗʰⁱˢ ʰᵃˢ ᵃ ᵈᵉᶜⁱᵐᵃˡ ᵖᵒⁱⁿᵗ ᵃˡʳᵉᵃᵈʸ.

ᵐᵒᵛᵉ ᵗʰᵉ ᵈᵉᶜⁱᵐᵃˡ ᵖᵒⁱⁿᵗ 6 ᵖˡᵃᶜᵉˢ ᵗᵒ ᵗʰᵉ ʳⁱᵍʰᵗ ᵗᵒ ᵍᵉᵗ ᵃ ⁿᵘᵐᵇᵉʳ ᵇᵉᵗʷᵉᵉⁿ 1 ᵃⁿᵈ 10 ʷʰⁱᶜʰ ⁱˢ 1.2

= 1.2 × 10⁻⁶

ᵗʰᵉ ⁱⁿᵈᵉˣ ⁱˢ ⁿᵉᵍᵃᵗⁱᵛᵉ ᵇᵉᶜᵃᵘˢᵉ 0.000 0012 ⁱˢ ᵃ ⁿᵘᵐᵇᵉʳ ˡᵉˢˢ ᵗʰᵃⁿ 1

3) 158.6

   

158.6 = 1.586 × 10²

(ᵐᵒᵛᵉ ᵗʰᵉ ᵈᵉᶜⁱᵐᵃˡ ᵖᵒⁱⁿᵗ 2 ᵖˡᵃᶜᵉˢ ᵗᵒ ᵗʰᵉ ˡᵉᶠᵗ ᵗᵒ ᵍᵉᵗ ᵃ ⁿᵘᵐᵇᵉʳ ᵇᵉᵗʷᵉᵉⁿ 1 ᵃⁿᵈ 10,ʷʰⁱᶜʰ ⁱˢ 1.586,ᵗʰᵉ ⁱⁿᵈᵉˣ ᵒʳ ᵖᵒʷᵉʳ ᶜᵒʳʳᵉˢᵖᵒⁿᵈˢ ᵗᵒ ᵗʰᵉ ⁿᵘᵐᵇᵉʳ ᵒᶠ ᵗⁱᵐᵉˢ ᵗʰᵉ ᵖᵒⁱⁿᵗ ʷᵃˢ ᵐᵒᵛᵉᵈ.)

4) 000.000589

    

 move the point 4 places to the right,the first two zeros are insignificant.

= 5.89 × 10⁻⁴

5)   If p = 5×10⁻² and q= 8× 10⁵,

      calculate p×q

solution:

p×q = 5×10⁻² × 8×10⁵

      = 5×8×10⁻²×10⁵

      = 40×10⁽⁻²⁺⁵⁾

(ᶠʳᵒᵐ ᵗʰᵉ ᵏⁿᵒʷˡᵉᵈᵍᵉ ᵒᶠ ˡᵃʷ ᵒᶠ ⁱⁿᵈⁱᶜᵉˢ)

 = 40×10³

 = 4×10¹ ×10³

 = 4×10⁴

(40 has to be converted to standard form because it is not a number between 1 and 10)

6) write 1.15 ×10⁻⁶ as a decimal.

      

solution:

     1.15 × 10⁻⁶

=   1.15 × 1

             -----

              10⁶

ᶠʳᵒᵐ ᵗʰᵉ ˡᵃʷ ᵒᶠ ⁱⁿᵈⁱᶜᵉˢ, (a⁻ⁿ = 1/aⁿ)

= 1.15

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

1 000 000

= 0.000 00115

7) calculate 7,200,000 ÷ 800,

ˡᵉᵃᵛⁱⁿᵍ ʸᵒᵘʳ ᵃⁿˢʷᵉʳ ⁱⁿ ˢᵗᵃⁿᵈᵃʳᵈ ᶠᵒʳᵐ

Solution:

write all in standard form.

= 7.2×10⁶ ÷ (8×10²)

= (7.2 ÷8)×(10⁶÷10²

= 0.9 × 10⁽⁶⁻²⁾

= 0.9 × 10⁴

(ᶜʰᵃⁿᵍᵉ 0.9 ᵗᵒ ˢᵗᵃⁿᵈᵃʳᵈ ᶠᵒʳᵐ ᵇᵉᶜᵃᵘˢᵉ ⁱᵗ ⁱˢ ⁿᵒᵗ ᵇᵉᵗʷᵉᵉⁿ 1 ᵃⁿᵈ 10)

= 9 × 10⁻¹ × 10⁴

= 9 × 10⁽⁻¹⁺⁴⁾       (ˡᵃʷ ᵒᶠ ⁱⁿᵈⁱᶜᵉˢ)

= 9 × 10³

8) Evaluate 

   

(3 × 10⁴) × (7 × 10⁵),

leaving your answer in standard form.

solution:

   3 × 10⁴ × 7 × 10⁵

= 3×7 × 10⁴ × 10⁵

= 21 × 10⁹

(21 Is not in standard form,so we convert it to S.F)

= 2.1 × 10 ×10⁹

= 2.1 × 10¹⁰

9) Evaluate

  √[(8.1 × 10⁻⁶) ÷(2.25 × 10⁷)]

  

= √[(8.1÷2.25) ×(10⁻⁶ ÷ 10⁷)]

= √[3.6 × 10⁽⁻⁶⁻⁷⁾]

= √[3.6 × 10⁻¹³]

= √(3.6 × 10 × 10⁻¹⁴)

(ʷᵉ ⁿᵉᵉᵈ ᵗᵒ ᵐᵃᵏᵉ 3.6  ᵃ ᵖᵉʳᶠᵉᶜᵗ square ᵗᵒ ʳᵉᵐᵒᵛᵉ ᵗʰᵉ square root)

= √(36 × 10⁻¹⁴)

= √36 ×√(10⁻¹⁴)

= 6 × 10⁻⁷

( 10⁻⁷ × 10⁻⁷ = 10⁻¹⁴)

10) if c = 2×10⁶ and y= 6 × 10⁵

find w² if w² = cy ÷ (c-y),

to 2 significant figures.

 solution:

w² = cy ÷ (c-y)

    = 2×10⁶×6×10⁵

    ---------------------

     ( 2×10⁶) -(6×10⁵)

        

    = 2 ×6×10⁶×10⁵

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

      (2×10×10⁵)-(6×10⁵)

   = 12 × 10¹¹

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

   (20×10⁵)-(6×10⁵)

   = 12 ×10¹¹

 ---------------------

     10⁵ ( 20-6)

    = 12 ×10¹¹

  ---------------

      14 ×10⁵

   = 12      10¹¹

      ----- × -----

       14     10⁵

   = (12/14) × 10⁶

   w² = 0.857 × 10⁶

         = 0.86 ×10⁶ 

        to 2s.f

NUMBER BASE

 

Number system is made up of different digit(called symbol) that are applicable for computation within a system. THE HIGHEST DIGIT OF ANY NUMBER SYSTEM IS ONE LESS THAN THE BASE.

The following are the basic number system 

*BINARY SYSTEM*(Base 2) : symbol used are 0 and 1. It has place values that are power of 2

*Octal system"(Base 8): Symbol used are 0,1,2,3,4,5,6 and 7. It has a place value which has the power of 8

*Denary system* (Base 10): Symbol used are 0,1,2,3,4,5,6,7,8 and 9. It has a place value which has the power of 10

*Duodecimal system*(Base 12):. It has a place value which has the power of 12. Symbol used are 0,1,2,3,4,5,6,7,8,9,X,and A

Where X = 10 and A = 11

*Hexadecimal system*(Base 16):. It has place values which has the power of 16.

Symbol used are: 0,1,2,3,4,5,6,7,8,9,A, B,C,D,E,F

Where, A=10, B=11, C=12, D =13, E=14, F=15

*Conversion from other bases to base  10*

RULE: 

 *worked examples*:

convert the following to base 10

(1)32145(base six)

Solution:

By expansion,we have:

3×6⁴+2×6³+1×6²+4×6¹+5×6°

=3888+432+36+24+5

=4385(base 10)

Hope you understand!

2.) 100101(base2)

By expansion,

1×2⁵+0×2⁴+0×2³+1×

2²+0×2¹+1×2°

=32+0+0+4+0+1

=37(base 10)

B· *conversion from base 10 to other bases*

Example 3:

Convert 89₁₀ to a binary number (base 2)

Solution:

By long division,we didvide by the new base(which is 2)

   2√89

   2÷44 r (1)

   2÷22 r (0)

   2÷11 r (0)

   2÷5   r (1) ∆

   2÷2   r (1) |

   2 ÷1   r (0) |

        0  r (1)  

Following the arrow's direction we have:1011001₂(base2)

Example 4:

Convert 75₁₀ to an octal number(base 8)

Solution:

Divide by the new base:

75÷8=9 r 3

9 ÷8=1 r 1

1 ÷8=0 r 1

=113₈(base8)

C. Conversion from a number which is not in base ten to another number ( not denary)

*WHEN CONVERTING A NUMBER WHICH IS NOT IN BASE 10 TO A BASE WHICH IS NOT DENARY. YOU FIRST CONVERT TO BASE 10, THEN TO THE REQUIRED BASE.

Example 5:

Express 307₈(base8)

in binary.

Solution:

We're to convert from base 8 to base 2

Step 1: convert to base 10 by expansion.

= 3×8²+0×8¹+7×8°

=192+0+7

=199(base 10)

Step 2: convert 199₁₀(base10) 

to base 2 by long division.

Solution:

199÷2=99 r 1

99÷ 2=49 r 1

49÷ 2=24 r 1

24÷2= 12 r 0

12÷2=. 6 r 0

6÷ 2= 3 r 0

3÷ 2= 1 r 1

1 ÷2= 0 r 1

= 11000111₂(base2)

Example 6

Express 111111(base2) in octal.:

Solution:

Convert to base 10

=1×2^5+1×2⁴+1×2³+1×2²+1×2¹+1×2°

=32+16+8+4+2+1

=63(base10)

 Then convert 63 to base 8 by dividing by 8

= 63÷8=7 r 7

     7÷8=0 r 7

= 77(base 8)

Example 7:

Convert 14312(base5) to base 6

Solution:

Step 1:first convert it to base 10.

= 1×5⁴+4×5³+3×5²+

1×5¹+2×5°

=625+500+75+5+2

=1207(base10)

Step 2:Then convert 1207 to base 6 by continuous division by 6

=

1207÷6=201 r1

  201÷6=33 r3

    33÷6=5 r3

      5÷6=0 r 5

=5331(base 6)

(D)BI - DEXIMALS:

These are decimals in binary form (base 2)

Note-When converting any number to number in base 10 that include decimal fraction,we multiply each digit of the number by the corresponding power of the base it is expressed.

8. Convert 0.234₅ to base 10 

Solution- 

0.234₅ = 0 x 5⁰ + 2 x 5⁻¹ + 3 x 5⁻² + 4 x 5⁻³

= 0 + 2/5 + 3/25 + 4/125

= 69/125₁₀ or 0.552₁₀

Example 9:

Solution:

Convert 1001.11(base two) to base 10

=1×2³+1×2²+1×2¹+

1×2°+1×2^-¹ +1×2^-² 

(The powers for the numbers after the decimal point are negative because they are less than 1, i.e,1001.11=1001+0.11)

=8+0+0+1+½+1/4

=9+¾=9 ¾

Example 10

Convert 4et(base12) to base 10

Solution:

[e stands for 11 and 

  t stands for 10]

= 4×12²+e×12¹+t×12°

=4×144+11×12+10×1

=718(base 10)

 E. *Addition, subtraction, multiplication and division in number bases*.

Example 11:

Simplify: 136+425+243{all in base 7}

Solutions:

Step 1:convert all to base 10

={1×7²+3×7¹+6×7°}

+{4×7²+2×7¹+5×7°}

+{2×7²+4×7¹+3×7°}

=49+21+6+196+14+5+98+28+3

=420{base ten}

Step 2:convert 420 to base7 by division.

= 420÷7=60 r 0

     60÷7=8. r 4

       8÷7=1 r 1

       1÷7=0 r 1

=1140₇(base7)

  

Example 12:

Simplify: 324-135₆{base 6}

Solutions:

Step 1:convert all to

 base 10 by expansion

={3×6²+2×6¹+4×6°}

-{1×6²+3×6¹+5×6°}

={124}- {59}

=65₁₀ (base ten)

Step 2:convert 65₁₀ to base 6

=

   65÷6=10 r 5

   10÷6= 1. r 4

    1÷6= 0 r 1

= 145₆

Example 13:

Find the product of 235₆and 14₆

Solution:

Convert all to base ten

235six=2×6²+3×6¹+5×6°=95ten

14₆ =1×6¹+4×6°

=10₁₀

=95×10=950₁₀

Step 2:convert 950₁₀ to base six.

=950÷6=158 r 2

  158÷6=26. r 2

   26÷6= 4 r 2

     4÷6= 0. r 4

=4222₆

Example 14

Divide 10010001₂ by 101two.

Solution:

convert both to base ten and divide.Then change the result to base 2

By expansion,10010001two= 

145ten and 

101two=5ten.

= 145÷5=29ten

= 29÷2=14 r1

   14÷2=7 r0

    7÷2 =3 r1

    3÷2= 1 r1

    1÷2= 0 r1

=11101two

  Hope you understand visit Standard forml