Say that the speed, $v$, of an object is given by a lognormal distribution:
$$N(\mu,\sigma,\xi)=\frac{1}{\sigma\sqrt{2\pi}}\exp{\left(-\frac{(\xi-\mu)^2}{2\sigma^2}\right)}$$
where
$$\xi=\log_{10}(v)$$
I want to figure out what the probability density function (PDF) $f(v)$ is.
Now, I'm pretty sure that $f(v)\neq N(\xi)$ because $N(\xi)$ is dimensionless, whereas I figure that $f(v)$ has units of 1/speed, i.e. $\rm m^{-1}\ s$, because the probability of a speed in the infinitesimal interval $[v,\ v+dv]$ is $f(v)\ dv$ which must be dimensionless (and $dv$ has units of $\rm m\ s^{-1}$).
But my intuition tells me that the probability $f(v)\ dv$ must be equal to the probability of a $\xi$ in the interval $[\xi,\ \xi+d\xi]$, so
$$f(v)\ dv=N(\xi)\ d\xi$$
Differentiating $\xi$, we get
$$\frac{d\xi}{dv}=\frac{1}{\ln10}\frac{1}{v}$$
$$\Rightarrow d\xi=\frac{1}{\ln10}\frac{1}{v}\ dv$$
Thus I conclude the PDF is
$$f(v)=\frac{1}{\ln10}\frac{N(\xi)}{v}$$
Is my reasoning correct? This is the first time I've encountered the lognormal distribution, and it's been a while since I last did anything to do with probability densities/PDFs.
The first thing that one typically attempts is to compute the cumulative distribution function (CDF) of $V$ (I will use capital letters for the random variables, $V$ for $v$ and $X$ for $\xi$) defined by $F_V(v) = P(V\le v)$ and then find the probability density by differentiation: $f_V = F_V'$. Since the transformation $X=\log_{10}V$ is monotonically increasing (and in one dimension), you have a good chance of finding explicit expressions in the following way: \begin{align*} F_V(v) &= P(V\le v) = P(10^X\le v) = P(X\le \log_{10}v) = \int_{-\infty}^{\log_{10}v} f_X(x)\, \mathrm dx, \\ f_V(v) &= F_V'(v) = f_X(\log_{10}v) \frac{1}{x\ln(10)} \end{align*} (here I denoted the probability density of $X$ by $f_X$ and used the chain rule to compute $F_V'$).
This is the same result as you got, except that you have to replace $N(\xi)$ by $N(\log_{10}v)$ (since you want $f(v)$ to be a function of $v$).
Your reasoning is correct as well, but it is more an intuitive or "physicist" way of thinking rather than a rigorous derivation.