interpretation of odds ratio when independent variable is log transformed In logistic regression, we derive the odds ratio by exponentiating the coefficient (estimate). If one of my independent variables ($X_1$) was log transformed (natural log), how should I interpret the odds ratio in such cases?
For example:

*

*$X_1$ has an OR of 0.46 (95% C.I: 0.24, 0.93) for a binary outcome of $Y$.

*$X_1$ was log transformed (natural log)

In the above results, usually (if no log transformation was done) one would have interpreted as "for every one unit increase in $X_1$, the odds of outcome $Y$ would have increased by 0.46". But this variable was log transformed (natural log, base = 2.72) during the pre-processing stage of the model creation.
Any suggestions pls
 A: Let's look at what's going on behind the scenes. A logistic regression model (binomial GLM) looks like this:
$$
\log\left(\frac{\pi}{1-\pi}\right)=\beta_0+\beta_1X_1
$$
assuming we're only fitting an intercept and one parameter for your $X_1$ variable.
As you correctly state, exponentiating the coefficients gives us the odds:
$$
\exp\left[\log\left(\frac{\pi}{1-\pi}\right)\right]=\exp(\beta_0+\beta_1X_1)\\
\implies \frac{\pi}{1-\pi} = \exp(\beta_0)\exp(\beta_1X_1)
$$
For simplicity, let's focus on $\beta_1$ and assume that $\beta_0=0$. This makes $\exp(\beta_0)=\exp(0)=1$, and therefore the above expression reduces to:
$$
\frac{\pi}{1-\pi} = \exp(\beta_1X_1)
$$
Now, let's look at what happens when $X_1$ increases by one unit e.g. from $2$ to $3$. Let's also imagine $\beta_1=-0.74$:
$$
(1) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times2)=\exp(-1.48)\approx0.23 \\
(2) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times3)=\exp(-2.22)\approx0.11
$$
A unit increase in $X_1$ has resulted in a decrease of $0.11/0.23=0.478$ in the odds. Note that this is just the odds ratio, and $\exp(\beta_1)=\exp(-0.74)=0.478$. So a unit increase in $X_1$ results in the odds of success being multiplied by the odds ratio, which is nothing but $\exp(\beta_1)$. Essentially, if $\beta_1>0$, an increase in $X_1$ will yield higher odds (because the odds ratio > 1), whereas if $\beta_1<0$, an increase in $X_1$ will yield lower odds (because the odds ratio < 1).
Now, you should be able to see for yourself what would happen if we were working with $\ln(X_1)$ rather than with $X_1$. Hint:
$$
(3) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(2))=2^{-0.74}\approx0.60\\
(4) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(3))=3^{-0.74}\approx0.44
$$
That is, a decrease in the odds of $0.44/0.60=0.74$. However, the odds ratio is still $\exp(\beta_1)=\exp(-0.74)=0.478$, so what's going on here?
By taking the natural logarithm of $X_1$, you change the interpretation of its coefficient. Now, the odds ratio represents the change in the odds if you multiply $X_1$ by $e$. Therefore:
$$
(5) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(2))=2^{-0.74}\approx0.599\\
(6) \space\space\frac{\pi}{1-\pi} = \exp(-0.74\times\ln(2\times e))=(2\times e)^{-0.74}\approx0.286
$$
We see that multiplying $X_1$ by $e$ results in the odds decreasing from $0.599$ to $0.286$. Here $0.286/0.599=0.478$ which is precisely the odds ratio i.e. $\exp(\beta_1)=\exp(-0.74)$.
Following the same logic, it's easy to see that:

*

*If we had transformed $X_1$ to be $\log_2(X_1)$, the interpretation would be "by how much would the odds decrease after doubling $X_1$"

*If we had transformed $X_1$ to be $\log_{10}(X_1)$, the interpretation would be "by how much would the odds decrease after multiplying $X_1$ by 10"

