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