Arithmetic with scientific notation
The final coefficient must be at least 1 and less than 10. Keep units and precision requirements in view
- Multiply coefficients and add indices Deal with the two parts separately \(\displaystyle (3\times10^4)(2\times10^{-6})=6\times10^{-2}\)
- Divide coefficients and subtract indices Subtract the entire denominator exponent, including its sign \(\displaystyle \frac{8\times10^{-3}}{4\times10^{-6}}=2\times10^3\)
- Normalise in the correct direction Dividing the coefficient by 10 increases the exponent by 1; multiplying it by 10 decreases the exponent by 1. \(\displaystyle 0.6\times10^3=6\times10^2\)
- Round only at the end Keep the exact value until rounding is requested. This example is rounded to two significant figures \(\displaystyle 1.764\times10^1\approx1.8\times10^1\)
Worked example
A sample contains 4.8 × 10⁶ grains with a mean mass of 3 × 10⁻⁶ mg per grain. Find the total mass to two significant figures
- \(\displaystyle 4.8\times3=14.4\)Multiply the coefficients
- \(\displaystyle 6+(-6)=0\)Add the signed exponents
- \(\displaystyle 14.4\times10^0=1.44\times10^1\)Normalise the exact result
- \(\displaystyle 1.44\times10^1\approx1.4\times10^1\text{ mg}\)Round once at the end to two significant figures
Common mistakes
Difficulty levels
Questions about this topic
Why not round the coefficients first?+
Early rounding can change the final answer. Retain the exact intermediate values and round at the end
What happens when rounding gives a coefficient of 10?+
Increase the exponent and use coefficient 1 with enough trailing zeros to show the requested precision
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 9 Mathematics, chapter 6: Indices and surds