Annoying math problem

[Sar]

Thanks a lot Urwumpe, it's not like I wanted to go to sleep or anything. :uhh: But this looks like something I could actually manage to do with my (still) a bit limited math skills. Might actually be nice to learn something outside the course frame for a change.

---------- Post added at 10:27 PM ---------- Previous post was at 07:47 PM ----------

Ok. it seems straight forward there's just few things I don't get in it!

[math] \sqrt{S} = 6 * 10^n [/math]

The n is made in this manner in the example

[math]D=6=2*2+2[/math]

D in the number of digits in the left of the decimal? That's what I understood from the rough estimate part, but after that I've got no real clue.

[math]D=8=2*2*2+2[/math] would be

[math] \sqrt{S} = 6 * 10^4 [/math] ?

if someone could write me an example or two I'd be really glad, I've got three hours of absolutely nothing coming up tomorrow which I could actually use with my pen and math scribble book.

Stoat: Thanks for the site! Seemed really usefull, I need to print those cheat sheets and check the site out properly.

And yes, I was sort of expecting to get buried in a pile of equations when I made the thread but so far this hasn't been too bad and I'm somewhat still on the same tracks that I was before, just further up

 

[Urwumpe]

Sar, you lost me there a bit, what are you trying to solve there?

also, you can start the calculations much easier by forming it into

[math]S = 36 \cdot 10^{2 n}[/math]

Remember:

[math](x^a)^b = x^{a \cdot b}[/math]

 

[Poscik]

Try [math]\sqrt{-4}[/math]

 

[fireballs619]

Try [math]\sqrt{-4}[/math]

No need to bring imaginary numbers into this now. If he's only just learning, this will only end up confusing him.

 

[Stoat]

I think it would be probably best to first take a step back and look at index notation and its rules. Then things like "imaginary" numbers and quadratic equations can be made much clearer when introduced.

 

[Sar]

I'll be looking at the index notation asap (tomorrow)

Anyway, forget the earlier post, I was way too tired.

Anyhoo, I spent three hours in a bar and among other nice things like seeing my friends I managed to figure (sort of) out the Heron's method.

[math]\sqrt{42} \approx 6[/math]
[math]((42 / 6)+6)/2 = 6.5[/math]
[math]((42 / 6,5)+6,5) / 2 = 6.4807[/math]

That was what I recalled from the method, the real one should be like this:

[math]1/2 * (6 + (42 / 6)) = 6,5[/math]
[math]1/2 * (6,5 + (42 / 6,5)) = 6,4807[/math]

I'd like to think that the end will justify the means so the sort of backwards way of calculating should be forgiven. It was rather nice to learn something cool for a change!

[math] \sqrt{-4} = -1*2[/math] or something. Not going there yet (or ever)

 

[Urwumpe]

Ok, you have two times written the same math in different orders...

[math] \frac{ \left ( \frac{42}{6} + 6 \right ) }{2} \equiv \frac{1}{2} \cdot \left ( 6 + \frac{42}{6} \right ) = 6.5 [/math]

 

[Sar]

yes it is, but it's somewhat simpler to make the equation in the order that's half times blah blah, that blah blah divided by two, but for some reason I did it in the latter way yesterday.

Anyway, I'll try to keep today and tomorrow to rewiev what I have learned so far and then it's off to fractional exponents!

 

[Sar]

Things I figured out I have learned in the first week I studied math in a some what serious way.

1. Elementary algebra, didn't really learn it since it was taught at the school way back when, but I now remember it again!
2. Learned to calculate percentages better
3. Learned more about exponents
4. Learned way more about roots (Thanks to this thread)
5. Heron's method
6. c1v1=c2v2

Actually the number 6 is something I'd like to ask about (sort of):

"You have 1 liter of 40% sugar/salt/alc/whatever solution and you need to make a 20% solution, but the dilutant (sp?) is 10% sugar/salt/alc/whatever"

This can't be calculated via c1v1=c2v2 method am I correct? How should I go about with this? For some reason I can't figure this one out, maybe it's the three hour sleep per night these past few days or just the fact that I haven't (yet) learned the tools to do so.

 

[Sar]

Ok. I have no idea what to do with this one, I know the answer is 3, but I have no idea on how to get that one out of this.

[math] ^x\sqrt{64}=4 [/math]

I know that this is true

[math] 4^x = 64 [/math]

but that doesn't really help me at all. Apparently ball is still a ball if you turn it around :/

I've tried everything I can think of, including randomly adding numbers together and dividing them.... Doesn't work.

 

[orb]

[math]^x\sqrt{64}=4[/math]

I'd change the root to exponent, and then the rest like solving exponential equations:

[math]^x\sqrt{64}=64^{1/x}[/math]

[math]4=64^{1/3}[/math]

[math]64^{1/x}=64^{1/3}[/math]

[math]1/x=1/3[/math]

[math]x=3[/math]