just reviewing two resources, I noticed a difference between the log normal p.d values :

One is here

enter image description here

which takes the e to the power in which it contains ln(x)

the other is here enter image description here

Which on page 5 , claims the same power with logarithm on the base of 10

which is correct ?

  • $\begingroup$ Please explicitly include formulas in your post: links dry up over time. It looks like your question concerns the arithmetic of working with logarithms, so in the meantime consider reviewing that. $\endgroup$
    – whuber
    Aug 2, 2018 at 15:17
  • $\begingroup$ I will do that when I get access to my PC, the rules of working with logarithms is clear, when you write ln you are indicating the basis is neper, if you write log, it's 10 unless specified $\endgroup$ Aug 2, 2018 at 15:41

1 Answer 1


In this context $\log$ means $\ln$ (i.e., a logarithm with base $e$), so the second website is probably not claiming that it is a logarithm with a base of ten. For a logarithm with base ten you would generally use the notation $\log_{10}$. The notation $\log$ is one of those nasty pieces of notation that is used to represent $\ln$ in some contexts and $\log_{10}$ in others, so you have to be able to spot which is meant from context.

Incidentally, the way I can tell that this is the case is that the density is wrong if this is a base ten logarithm. If you were to define a random variable $X = 10^Y$ with $Y \sim \text{N}(\mu, \sigma^2)$ then you have:

$$\Bigg| \frac{dy}{dx} \Bigg| = \Bigg| \frac{d}{dx} \log_{10}x \Bigg| = \frac{\log_{10} e}{x},$$

so the pdf of $X$ would be:

$$p_X(x) = \frac{\log_{10} e}{x \sigma \sqrt{2 \pi}} \cdot \exp \Big( - \frac{1}{2} \Big( \frac{\log_{10} x - \mu}{\sigma} \Big)^2 \Big).$$

So you can see that if you were to define a lognormal random variable on the base-10 scale then it would have an additional scaling term $\log_{10}e$ in order to make it integrate to one.


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