To begin with, $Y = -\ln(X)$ only works when $[a,b] \subset \mathbb{R}_{>0}$. So, let
$$
X \sim \mathcal{U}[a,b] \quad\text{where}\quad [a,b] \subset \mathbb{R}_{>0}.
$$
then $Y$ is log-uniformly distributed, in symbols
$$
Y \sim \mathcal{LU}[\alpha,\beta] \quad\text{with}\quad \alpha = -\exp(a) \ \text{and} \ \beta = -\exp(b).
$$
This can be derived by applying the transformation rule to the probability density function (PDF) of $X$, given by
$$
f_\mathcal{U}(x \mid a,b) = \begin{cases} \dfrac{1}{b - a}, & a \leq x \leq b \\ 0, & \text{else} \end{cases}.
$$
Suppose $\alpha = -\exp(a)$ and $\beta = -\exp(b)$. The PDF of $Y = -\ln(X)$ is then
$$
\begin{eqnarray}
f_\mathcal{U}(-\ln(x) \mid a,b) \cdot \left| \dfrac{\mathrm{d} (-\ln(x))}{\mathrm{d}x} \right|
&=& \begin{cases} \dfrac{1}{(b - a) \cdot x}, & a \leq -\ln(x) \leq b \\ 0, & \text{else} \end{cases} \\
&=& \begin{cases} \dfrac{1}{(b - a) \cdot x}, & -\exp(a) \leq x \leq -\exp(b) \\ 0, & \text{else} \end{cases} \\
&=& \begin{cases} \dfrac{1}{(\ln(-\beta) - \ln(-\alpha)) \cdot x}, & \alpha \leq x \leq \beta \\ 0, & \text{else} \end{cases} \\
&=& \begin{cases} \dfrac{1}{\ln\left( \dfrac{-\beta}{-\alpha} \right) \cdot x}, & \alpha \leq x \leq \beta \\ 0, & \text{else} \end{cases} \\
&=& \begin{cases} \dfrac{1}{(\ln(\beta) - \ln(\alpha)) \cdot x}, & \alpha \leq x \leq \beta \\ 0, & \text{else} \end{cases} \\
&=& f_\mathcal{LU}(x \mid \alpha,\beta).
\end{eqnarray}
$$
The result is the PDF of the log-uniformly distributed variable $Y \sim \mathcal{LU}[\alpha,\beta]$.