Operations with Standard Form (Edexcel IGCSE Maths)

Spec point — 1.9A

Operations with standard form

How do I perform calculations in standard form using a calculator?

• Make use of brackets around each number, and use the box x 10 ^ x button to enter numbers in standard form

  • e.g. $( 3 \times 10^{8 ) \times ( 2 \times 10^{-3} } )$
  • You can instead use the standard multiplication and index buttons

• If your calculator answer is not in standard form, but the question requires it:

  • Either rewrite it using the standard process

    • e.g. 3 820 000 = 3.82 × 10⁶

  • Or rewrite numbers in standard form, then apply the laws of indices

    • e.g. 243 × 10²⁰ = (2.43 × 10²) × 10²⁰ = 2.43 × 10²²

Exam Hint Exam questions tend to use large numbers as indices. This is so that the question can test whether you know how to calculate using standard form.

How do I multiply and divide numbers in standard form using index laws?

Multiplication

• Consider the example $( 3 \times 10^{112} ) \times ( 4 \times 10^{235} )$

• STEP 1 Multiply the ordinary numbers

  • e.g. $3\times4=12$

• STEP 2 Multiply the powers of 10 by adding the indices

  • e.g. $10^{112}\times10^{235}=10^{112+235}=10^{347}$

• STEP 3 Multiply the new ordinary number and the new power of 10

  • e.g. $12\times10^{347}$

• STEP 4 (if needed) Write the ordinary number in standard form and simplify the powers of 10 by adding the indices

  • e.g. $12=1.2\times10^{1}$ and $1.2\times10^{1}\times10^{347}=1.2\times10^{348}$

Division

• Consider the example $( 2 \times 10^{- 150} ) \div ( 8 \times 10^{- 131} )$

• STEP 1 Divide the first ordinary number by the second ordinary number

  • e.g. $2\div8=0.25$

• STEP 2 Divide the first power of 10 by the second power of 10 by subtracting the indices

  • e.g. $10^{- 150} \div 10^{- 131} = 10^{- 150 - ( - 131 )} = 10^{- 19}$

    • Be careful with negatives!

• STEP 3 Multiply the new ordinary number and the new power of 10

  • e.g. $0.25\times10^{-19}$

• STEP 4 (if needed) Write the ordinary number in standard form and simplify the powers of 10 by adding the indices

  • e.g. $0.25=2.5\times10^{-1}$ and $2.5\times10^{-1}\times10^{-19}=2.5\times10^{-20}$

How do I add and subtract numbers in standard form using index laws?

• Consider the example $( 4 \times 10^{150} ) - ( 2 \times 10^{148} )$

• STEP 1 Rewrite the number with the biggest power of 10 so that it has the same power of 10 as the number with the lowest power of 10

  • e.g. $4\times10^{150}=4\times10^{2}\times10^{148}=400\times10^{148}$

    • The ordinary gets bigger as the power of 10 gets smaller

• STEP 2 Collect like terms by adding or subtracting the ordinary numbers

  • e.g. $( 400 \times 10^{148} ) - ( 2 \times 10^{148} ) = 398 \times 10^{148}$

    • Do not change the power of 10

• STEP 3 Write the ordinary number in standard form and simplify the powers of 10 by adding the indices

  • e.g. $398=3.98\times10^{2}$ and $3.98\times10^{2}\times10^{148}=3.98\times10^{150}$

• This method works for negative powers too

  • e.g. consider $( 8 \times 10^{- 120} ) - ( 5 \times 10^{- 121} )$

    • $8\times10^{-120}=8\times10^{1}\times10^{-121}=80\times10^{-121}$
    • $( 80 \times 10^{- 121} ) - ( 5 \times 10^{- 121} ) = 75 \times 10^{- 121}$
    • $75=7.5\times10^{1}$ and $7.5\times10^{1}\times10^{-121}=10^{-120}$

Worked Example Show how $( 45 \times 10^{- 123} ) \div ( 0.9 \times 10^{105} )$ can be written in the form $A\times10^{n}$, where $1\leq A<10$ and $n$ is an integer.

Answer:

Rewrite the division as a fraction, then separate out the powers of 10

$\frac{45\times10^{-123}}{0.9\times10^{105}}=\frac{45}{0.9}\times\frac{10^{-123}}{10^{105}}$

Work out $\frac{45}{0.9}$

$\frac{45}{0.9}=\frac{450}{9}=50$

Work out $\frac{10^{-123}}{10^{105}}$ using laws of indices

$\frac{10^{-123}}{10^{105}}=10^{-123-105}=10^{-228}$

Combine back together

$( 45 \times 10^{- 123} ) \div ( 0.9 \times 10^{105 ) = 50 \times 10^{-228} }$

Rewrite in standard form, where $a$ is between 1 and 10

$50\times10^{-8}=5\times10^{1}\times10^{-228}=5\times10^{-227}$

$5\times10^{-227}$

Worked Example Given that $a\times b=c$, where

$a=5\times10^{20}$ $b=6\times10^{n}$ $c=3\times10^{33}$

Find the value of $n$.

Answer:

Substitute the values into the given equation, $a\times b=c$

$5\times10^{20}\times6\times10^{n}=3\times10^{33}$

Rearrange so the powers of 10 are grouped together

$5\times6\times10^{20}\times10^{n}=3\times10^{33}$

Calculate the 'number' part and use laws of indices to combine the powers of 10

$30\times10^{20+n}=3\times10^{33}$

The two sides of the equation are almost in the same format, but we need to change the 30 to a 3

Dividing the number by 10, means the power of 10 must be increased by 1 to compensate

$3\times10^{21+n}=3\times10^{33}$

The two sides of the equation now match, so the powers of 10 on each side must be equal

$21+n=33$

Subtract 21 from both sides to find $n$

$n=12$