Alexander,
No, the code is correct. Let x = log_10 p. Then
e^(x ln 10) = 10^x = p.
> inline Prob LogPtoProb(LogP2 prob)
> {
> if (prob == LogP_Zero) {
> return 0;
> } else {
> return exp(prob * M_LN10);
> }
> }
&
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