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NASA now has a satellite orbiting Mercury and snapping photos. http://blog.seattlepi.com/aerospace/2011/03/29/heres-mercury/ http://www.latimes.com/news/science/la-sci-messenger-20110330,0,7061787.story The second article is from the LA Times. It states that NASA's photos capture "areas near the planet's southern pole that have never been seen before — areas that could host water in the form of ice." How is this even possible? The surface temperature of Mercury would instantly evaporate any form of water, without even taking into account the amount of pressure that close to the sun.

I think part of the argument against you is that there are natural replacements for fertilizer (high in nitrogen) that are cheaper than fertilizer. I'm sure artificial fertilizers have more bang for their buck (efficiency) pound for pound, but the cost of that pound is also higher than an organic alternative. For example: using a cod as a nutrient base for a plant as opposed to an artificial fertilizer. Let's say they both work well. The artificial fertilizer is more efficient for the, say, 10 grams used. If you used an equivalent mass of cod, the same results would not be achieved. My argument here is that the efficiency of a product should be termed as a product of initial cost and total output instead of a product of mass and output. The higher the difference between the former two, the higher the efficiency.

Hi, barring a pre-pharm program, what undergraduate major is best suited to getting a high gpa whilst still obtaining the required courses for pharmacy school? (in canada)

Erm... rainfall will leach urea and such from manure. which is good. ?

ah forgot about the squaring. gotcha.

Yeah, sorry, lol. Okay, so for that example, if we assign some "x" values for explanation purposes... for example, in n=1, the node in the middle can be x=2. for n=2, the peak of of the wave is at x=1 and the trough is at x=3 in n=1, its obvious from the peak amplitude that the probability of finding the particle is highest in the centre of the box (x=2) but in n=2, wouldn't the probability be lowest in the trough (x=3) and highest at the peak (x=1)?

yeah, sorry if i snapped.. it was like 6am and I'd been up all night typing out equations. I'll hit up the library, thanks

For n=2, wouldn't the lowest probability be where the wavelength is lowest?

If I could afford a 200$ textbook I'd just get tutoring and wouldn't be here. Thanks for everything.

can you link me a guide on that? i'm not really sure where to start. something pertinant to the example would be helpful

thatt seems proper. why would wolfram show me that though? is it the exponential equivalent of imaginary numbers?

an asymptote is a value that cannot exist and other values approach that limit, right? sure, i get that, but lets forget about the asymptote itself. how can this: [math] y=log(-x) [/math] be valid?

sorry bout that, now that i know how to use latex it'll be easier to express equations from now on any ideas on how to get values from log(-1)? that seems to me to be an invalid domain. yet wolfram yields results

oh, im sorry. i was working on another problem and i must have mixed up the two pages. the wolfram page you posted is the correct one... and yeah, haha, i know it is a bit of a jump. but i realized the answer to my op and instead of creating a new thread for another question, i just decided to put it here instead. [math] y = -2log_3(x-3) -1 [/math] (thanks, cap'n) so how can x be less than three? in other words, i know the vertical asymptote isn't crossed, but how can there be a log(-1)?

y = -2log(base3)(x-3) -1 here's the wolfram page. http://www.wolframalpha.com/input/?i=%283%29%28log%289x-36%29%29%2Flog%283%29%29-13+ the negative one is the vertical shift of one downwards, not an exponent (incidentally, how do you type like that?)

------ okay, so that was something i could have figured out if i wasn't dead tired =p i've got it now. but here's something. y=-2log_3(x-3)-1 I've graphed this out and there seems to be an asymptote at x=3. Intuitively, I can say that x cannot be less than 3. HOWEVER. When I graph with a calculator, it shows values at x=2 and below. How can this be? (eg coordinates= (2, -1), (0, -3)

So a wave function is a function of probability and position for a specific point in time (eg. t=0.5 seconds)? Can someone post a picture of a wave function with explanations of its properties?

sorry, dumb question.

Thanks; real-life examples are very, very helpful in explaining physical phenomena. Especially those quantum quirks that seem to elude my understanding =)

My (vague) understanding is that when a determined particle (in this case, a photon), interacts with an electron probability cloud (wave function), the electron "pops" into existance at the point of interaction. I have no idea why this is.

My thinking behind this is that any environmental "noise" will cause a wave function to collapse upon interaction with the particle. (At least in my understanding.) Maybe I should be more specific. Is EM radiation from measuring devices/humans the cause of wave function collapse in the double slit experiment (relating to the observer effect)? Another way of asking this is: Is the observer effect attributable to EM radiation?

Ah, ok. I was looking at this the wrong way. I was using "1" as the "x" value in log_b(x) = y It should have been the "y" value, as a log equation would result in "1" So in: log_9(x) = 1 9^1=x x=9 Starting Over: log_9((x-6)(x+2)) = 1 log_9((x-6)(x+2)) = log_9(9) (x-6)(x+2) = 9 x^2 - 4x - 12 - 9 = 0 x^2 - 4x - 21 = 0 Simple trinomial factoring: (x-7)(x+3) = 0 x= 7 OR x= -3 (undefined) Therefore, x = 7 Checking Answer: log_9(7-6) + log_9(7+2) = 1 log_9(1) + log_9(9) = 1 log_9(1)(9) = 1 log_9(9) = 1 In exponential form: 9^1 = 9 Sweet. Nothing feels as good as FINALLY solving a math problem. xp+10 (chugs beer) Thanks for the help, Mooey. Haha, I just realized I called you "money" before... just another example of me rushing over things and my brain filling in the gaps. I have trouble taking things slow, which is why I take the pains to check my answers so many times. Its saved me more than once.

Hey, thanks money, that was a stupid error, haha. The thing is, I see now where I skipped a step, and I'm not sure how it led me to the right answer. Well, the answer's correct now, but I see what you mean about skipping steps. Let's go back to the beginning, so that I actually completely understand what I am doing. We have: log_9(x-6) + log_9(x+2) = 1 The next step is to convert the 1 into something with the base 9 that results in a value of "1" Hmm, what could that be? log_9(x-6) + log_9(x+2) = 9^0 And in log form.. log_9(x-6) + log_9(x+2) = log_9(1) Let's apply the laws of logs. log_9( (x-6)(x+2) ) = log_9(1) Here's where I'm stuck, again. I want to get to: (x-6)(x+2) = 0 But doing so would require me to solve the log on one side of the equation, while dropping the log on the other side. Which doesn't seem right. What seems right is: log_9( (x-6)(x+2) ) = log_9(1) (x-6)(x+2) = 1 This seems correct because in the form "log_b(y) = z," the "log_b" is dropped, and they "y" value is kept Which goes to: x^2 - 4x -12 -1 = 0 x^2 - 4x - 13 = 0 And in the quadratic equation... x = (-b +- (b^2 - 4ac)^(.5)) / 2a x = (4 +- (16 - 52)^(.5)) / 2a x = (4 +- 6i) / 2 x = 2 +- 3i So in this method: x = 2 + 3i OR x = 2 - 3i I'm not sure how to check if these are correct answers, since they are complex numbers, which I have little experience with [other than knowing "i=(-1)^(.5)"].

Ahhhh, the quadratic formula: x = -b +- ( (b^2 - 4ac)^(.5) )/2a I always seem to overlook this lil equation because its just so darned simple to use. I immediately go for the hardest solutions XD Anyways, onwards: x = 4 +- ( (16 + 48)^(.5) ) /2 x = 4 +- ( (64)^(.5) ) /2 x = 4 +- 8 / 2 x = 4 +- 4 x = 8 OR x = 0 (undefined) Therefore, x = 8 Ugh, ok I just plugged that Into the original formula and it is wrong.... x=8 yields ~1.363416514 instead of ~1 and in x^2 - 4x -12 = 0 64 - 32 - 12 = 0 32 - 12 = 0 20 = 0 As you might already know, twenty does NOT equal zero. Something is very wrong. I can't seem to figure out any way to factor this: x^2 - 4x -12 = 0 I don't think it can be factored in a useful way, can it?

As far as I remember, Log(baseY)x + log(baseY)2x = log(baseY)(x(2x)) But I've already done this: log(base9)(x-6) + log(base9)(x+2) = 1 log_9((x-6)(x+2)) = 1 log_9((x-6)(x+2)) = log_9(0) log_9(x^2 - 4x - 12) = log_9(0) x^2 - 4x = 12 I'm not sure about this next part. 2logx - log4x = log12 2logx = log(12) + log(4x) 2logx = log(12)(4x) 2logx / log4x = log12 log12 = 2logx / log4x log12 = log((x^2)-(4x)) I'm going in circles here. 12 = x^2 - 4x I'm still stuck. Could I have another hint please? Sorry, I still don't get it. If I was to turn it into a logarithm with a base of 9, it would be: log_9(1)=0 The only number I can think of that'll apply the Log Rule of Multiplication is 1. log_9(1) + log_9(1) = 0 log_9(1)(1) = 0 Seems kind of useless.. I'm missing something here.