# How can logistic loss return 1 for x = 0?

I have looked at the logistic loss function at many different sources, and many places I find it plotted like shown here:

Taken from http://fa.bianp.net/blog/2013/loss-functions-for-ordinal-regression/

The logistic loss is typically defined as $L(y_i, f(x_i)) = log(1+exp(x_i y_i))$ where $y_i = \pm \;1$.

What confuses me is that when calculating the logistic loss for $x_i = 0$, I cannot see how to get a value of $1$ when inserting into the formula:

$L(1, f(0)) = L(-1, f(0)) = log(1+exp(0)) \approx 0.69$

Have I misunderstood the concept? I can only see that a loss of $1$ can be returned by using a logarithm with base $1$, but I cannot imagine that would be anyone's intention.

A logarithm with base 2 would give $log_2(1 + exp(0)) = log_2(1+1) = log_2(2) = 1$.
Logarithms aren't defined for base 1; you can see how it isn't really convenient to define $log_1$ by considering the formula for change of base: $log_b(x) = \frac{log_k(x)}{log_k(b)}$. If we were trying for $log_1$ we'd get $log_k(1) = 0$ in the denominator, which is inconvenient. Or consider that we're trying to find a number $p$ so that $1^p = x$, which is only possible for $x=1$, and then you could use any $p$. See this Q and also this Q on the math forums.