You're right that the quadratic term will, eventually, dominate. However, for the range I considered (0 < XP < 100), adding that term results in a slightly better fit.

But, the fit is nearly as good without it, and so for interpretive purposes (instead of get-the-best-fit purposes), dropping the quadratic term makes for a better model:

Read 99 items
Read 99 items

Call:
lm(formula = log10(count) ~ xp)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.16823 -0.10095 -0.01733  0.07757  0.25553 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept)  4.194851   0.023379  179.43   <2e-16 ***
xp          -0.027268   0.000406  -67.17   <2e-16 ***
---
Signif. codes:  0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 

Residual standard error: 0.1154 on 97 degrees of freedom
Multiple R-Squared: 0.979,    Adjusted R-squared: 0.9787 
F-statistic:  4512 on 1 and 97 DF,  p-value: < 2.2e-16
[download]

With this model, our estimating function is as follows:

sub estimate_count_from_xp($) {
    my $xp = shift;
    10 ** ( 4.195 - 0.2727 * $xp );
}
[download]

From this, it's easy to see that we have classic exponential decay w.r.t. XP.

Does this match your intuition?