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
