<p>We have that:
<span class="math-container">$$
\ln(\text{pop in year 2}) - \ln(\text{pop in year 1}) = \ln\left(\frac{\text{pop in year 2}}{\text{pop in year 1}}\right)
$$</span>
So if both are measured in millions or in thousands, the resulting number will be the same. In other words, if you take log differences, the scale of measurement does not matter.</p>

CC BY-SA 4.0
2022-06-14T07:33:38.100Z
2022-06-14T07:33:38.100Z
tdm
<p>I want to approximate annual population growth using logarithms; that is, log(population at year 2) - log(population at year 1). Is it okay to use population in thousands, or in millions in these approximations? For example, Maddison Project and PWT provide the population data in thousands and millions.</p>

CC BY-SA 4.0
2022-06-10T10:41:37.830Z
By "in millions", do you mean expressed in millions (eg 24.723 M), or rounded to the nearest million (eg 25 M)?
Adam Bailey
CC BY-SA 4.0
2022-06-11T06:27:15.740Z
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
Giskard
<economic-growth>
Calculating population growth in logs
Ornithogalum
# Calculating population growth in logs

(Asked by: username_0 on 2022-06-09)

I want to approximate annual population growth using logarithms; that is, log(population at year 2) - log(population at year 1). Is it okay to use population in thousands, or in millions in these approximations? For example, Maddison Project and PWT provide the population data in thousands and millions.

**Tags:** economic-growth
> **Comments:**
> **username_1**: By "in millions", do you mean expressed in millions (eg 24.723 M), or rounded to the nearest million (eg 25 M)?  
> **username_2**: Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.  

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## 1 answers

### Answer by username_3 on 2022-06-14. Score: 3

We have that:
$$
\ln(\text{pop in year 2}) - \ln(\text{pop in year 1}) = \ln\left(\frac{\text{pop in year 2}}{\text{pop in year 1}}\right)
$$
So if both are measured in millions or in thousands, the resulting number will be the same. In other words, if you take log differences, the scale of measurement does not matter.


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