Recent content by Char. Limit

I'll try for six years presence to make up for six years absence, how's that sound? :wink:

Thank you! And thanks for the welcome from @WWGD and @Borg as well. It's been a good while but I'm glad to see you're doing well.

What if I popped up here again? Wouldn't that be weird?

Hillary Clinton, with 291 electoral votes and, say, 48.8% of the popular vote. That's my guess. I'm assuming she'll win Nevada, New Hampshire, and North Carolina, but lose Florida. I think that's not too bad an assumption.

For now, at least. I'm not dead, which I'm sure you're happy to hear. Let's hope my presence can last. It's Friday here too. My co-worker has been absent for two and a half days in a row now, but the workweek is over and I can relax now. So all in all, pretty nice. Hell is coming though...

Hey there. How's it going?

Yes it does. That follows from the definition of the square root. I would explain in further detail, but I don't have the background knowledge for that myself. Hopefully another member will.

Because it does. That's how i is defined, it's the complex number defined such that i^2 = -1. In general, the rule of sqrt(a)*sqrt(b) = sqrt(a*b) only works when a and b are both positive reals. That's not the case here.

Huh. That's actually not any trig I ever learned... very good to know, though. My bad, pas, that's definitely the way to go.

Honestly, this doesn't look easy in any way, if I'm completely honest. I did some checking on Wolfram-Alpha to see what needs to be done, and it looks like it needs to be somehow written in terms of the tangent function... which I suppose you could do by dividing the whole thing by cos(2x) and...

I replaced c^2 with various small square numbers, just to see if there was any sort of pattern. There... wasn't one, though it was interesting that 4 and 9 generated a result using only elliptic integrals of the first kind, while anything above that started throwing the imaginary unit and...

Ah, thank you, that's the one! Of course, the exponential function is only one-to-one if we assume x and y are both real, but we can safely assume that here.

And that right there, that's the assumption. We assume that x is close to a - namely, that |x - a| < 1. We can safely assume this, so we do, and the rest of the proof follows from such.

I think I see what's going on here. hddd, your problem is that you're assuming all definitions that involve definite integrals have the lower endpoint of the integral starting at zero. From what I can tell, you're working off the following definition: ln(x) \colon = \int_{0}^{x} \frac{1}{t} dt...

Okay, let's see. So just to make sure I have the problem clear... your two equations are 2^{x+y} = 6^{y} and 3^{x} = 6 \cdot 2^{y}. Fun looking problem, really. Well, let's see... if you multiply both sides of the second equation by 2^x, you get this: 6^{x} = 6 \cdot 2^{x+y} Then if we...