Simplifying ln(e^(x) + 1)

Bah.

I haven’t had calc for like… 2 semesters, and a friend asked me to check over his work to see if he had made some eggregious error. I searched Google real quick, and my log rules are fuzzy, at best.

Is there a way to simplify ln(e^(x) + 1) any further? I’m leaning heavily towards no.

No. I was going to say you could make an approximation near x=0 that’s simpler, but I don’t even think that’s the case.

No, there isn’t. You can write it in a couple of other equivalent forms, but I wouldn’t call them simpler.

I don’t think that there’s any way ln(e[sup]x[/sup]+1) can be simplified.

Near x=0, I suppose you could expand e[sup]x[/sup], in terms of a series expansion, and then discard higher order terms, and then expand the resulting log, again in terms of a power series.

But I can’t see any reason why one would want to do that. Why do you want to simplify it?

The most you can say is that ln(e[sup]x[/sup] + 1) is approximately equal to x, with the approximation being better for larger values of x.

Actually, at the other end of things, ln(exp(x) + 1) = exp(x) is good to better than 1% for x < -4.

Did you double check to make sure it wasn’t really supposed to be
ln(e^(x+1))? That, of course, you could simplify.

And! The graph of this function can be expressed parametrically as:

x = 2ln(sinh(t))
y = 2ln(cosh(t))

That’s pretty simple, huh?