Energy and Human Ambitions on a Finite Planet, 2021a

A.7 Scientific Notation 367
More generally, in Chapter 1, we saw that we can represent any base, b,
raised to some arbitrary number, n, as:
b n e n·ln b 10 n·log 10 b ,
(A.11)
where we use the exponential function <strong>and</strong> its inverse function (natural
log, ln), or alternatively the base-10 equivalents. If, for some reason, we
lacked a y x calculator button, these approaches allow more fundamental
ways to get at the same thing.
A.7 Scientific Notation
The single-biggest mistake students make when it comes to scientific
notation is easily remedied by underst<strong>and</strong>ing it not as a set of rules, but
for what it’s actually doing.
Most of the time, students get it right: they see 1.6 × 10 2 <strong>and</strong> move the
decimal to the right two times to get 160. A little harder is negative
exponents, like 2.4 × 10 −2 . Moving the decimal point twice to the left
results in the correct 0.024 answer.
The hangup can come about if the process is misconstrued as simply
“counting zeros.” Ironically, a student might correctly convert 6 × 10 3
by adding three zeros to the 6 to get 6,000, but then mistake 10 3 for
10,000—thinking: start with 10 <strong>and</strong> add three zeros.
The sure-fire way is to connect to the concept of integer powers, so that
10 3 is simply 10 · 10 · 10, which is unmistakably 1,000. Likewise, 10 −4 is
four repeated (multiplied) instances of 10 −1 , each one representing 1
or 0.1. String four together, <strong>and</strong> we have
the basics.
1
10,000
10 ,
, or 0.0001. So fall back on
Example A.7.1 We can also apply the rule of multiplying exponentiated
quantities covered in Eq. A.8. So3.2 × 10 3 times 2 × 10 2 can be
written out as 3.2 · 2 · 10 3 · 10 2 (order does not matter), which we can
recognize as 6.4 × 10 5 .
What about division: 2.4 × 10 13 divided by 8 × 10 7 ? Several ways to
approach this might be instructive. Let’s ignore the pre-factors (2.4 <strong>and</strong>
8) for now <strong>and</strong> focus on the powers of ten. The st<strong>and</strong>ard practice is to
subtract the exponent in the denominator from that in the numerator:
13 − 7 6 in this case, so that we are left with 2.4
8 × 106 . We could also
represent the 10 7 in the denominator as 10 −7 in the numerator, as per
Eq. A.9. Now we just add the exponents to get the same result. Or we
could invert the 8 × 10 7 to become 0.125 × 10 −7 <strong>and</strong> multiply this by
2.4 × 10 13 .
But I want to present the way I would do it to make it easy enough
© 2021 T. W. Murphy, Jr.; Creative Commons Attribution-NonCommercial 4.0 International Lic.;
Freely available at: https://escholarship.org/uc/energy_ambitions.