1.1 Scientific notationIB Maths: Analysis and Approaches SL: Revision notes

A number is in scientific notation (also called standard form) when it is written as $a\times10^{k}, \quad 1\le a<10,\ k\in\mathbb{Z}.$ The integer $k$ is the exponent. So $4.35\times10^{12}$ is in scientific notation, but $43.5\times10^{11}$ and $0.435\times10^{13}$ are not, even though they are equal to it.

• Large numbers have a positive exponent: $86\,400 = 8.64\times10^{4}$.
• Small numbers have a negative exponent: $0.000\,052 = 5.2\times10^{-5}$.
• Numbers from 1 up to 10 have exponent 0: $7.1 = 7.1\times10^{0}$.

Deal with the front numbers and the powers of 10 separately, using the laws of exponents: $(a\times10^{m})(b\times10^{n}) = ab\times10^{m+n}, \qquad \frac{a\times10^{m}}{b\times10^{n}} = \frac{a}{b}\times10^{m-n}.$ Then adjust so that the front number is between 1 and 10. For example $(8\times10^{-6})(2.5\times10^{13}) = 20\times10^{7} = 2\times10^{8}.$ Dividing: $\frac{1.5\times10^{11}}{3\times10^{8}} = 0.5\times10^{3} = 5\times10^{2}.$ When the front number is multiplied by 10 the exponent goes down by 1, and vice versa, so the value is unchanged.

You can only add or subtract the front numbers when the powers of 10 are the same. Rewrite one number so that both share a power of 10 (usually the larger one), then combine: $4.5\times10^{12} - 1.5\times10^{11} = 4.5\times10^{12} - 0.15\times10^{12} = 4.35\times10^{12}.$ If the exponents differ a lot, the smaller number barely changes the answer — a useful check. For example $3.2\times10^{11} + 1.05\times10^{13} = 1.082\times10^{13}$, only slightly more than $1.05\times10^{13}$.

A calculator may display 5.2E30. This is not acceptable in an IB answer: you must write $5.2\times10^{30}$.

Unless told otherwise, give answers exactly or to 3 significant figures. In scientific notation the significant figures are the digits of $a$: $7.39\times10^{3}$ has 3 s.f.

In real-world problems, check the answer makes sense in context. If a data centre can hold $8\times10^{10}$ photos and $1.5\times10^{8}$ arrive per day, $533.3$ days means it first becomes full during day 534 — round according to the situation, not automatically.