Math ..

[Dendrek]

I'll be around.

[Kezotar]

Not a problem. I should be checking gen OT periodically for the next hour or two.

I'll be around.

Sweet guys!

[Kezotar]

Okey, so I'm kinda having a hard time, I'm at this..

I've managed to crumble these together, however I just can't get to find x on nr 1 and n on the nr 2. The thing is, I have a problem basically doing it, I'm not sure how I should go on, step wise.. If anyone might have a hint..

[haxartus]

What exactly is the problem ? You can't do the equations ?
In the first one you distribute the ^20 both denominator and numerator and multiply everything by the expression before the parentheses. Numerator becomes x (because 1^20 is just 1), denominator is 1.04^21. The other number is jut x/1.04. When you divide fractions you just multiply by the opposite.
The second one is very similar.

[Kezotar]

What exactly is the problem ? You can't do the equations ?
In the first one you distribute the ^20 both denominator and numerator and multiply everything by the expression before the parentheses. Numerator becomes x (because 1^20 is just 1), denominator is 1.04^21. When you divide fractions you just multiply by the opposite.
The second one is very similar.

Yes, I can't do the equation..

If it helps, I'm using a calculator TI 84 +.

Oh, I didn't quite understand

[Blueobelisk]

Just start plugging in the numbers and writing the decimals in place then.

Biggest step to doing it by hand is identifying the main operator on the left, and undoing that operation until you get what you want. For example: (5*x)/2 = 10. First get rid of the two on the denominator by multiplying both sides by two, then get rid of multiplication by dividing by 5.

To be honest this isn't really a technique problem. It's more of a...know algebra problem.

[Kezotar]

I think I managed

(1)


What I did was taking ((1/1,04^20)-1)/((1/1,04)-1) on the calculator.

I had then x/1,04 * 14,13 = 1.500 000.

I then took, 1.500 000/14,13

x/1,04 = 106 157

x = 106 157 * 1,04..

x = 110403

However the answer is 110373 but it's probably because I took bits for bits rather than mashing the whole thing in the calculator.

[Blueobelisk]

Write it all in one step on your calculator then.

[Kezotar]

Just start plugging in the numbers and writing the decimals in place then.

Biggest step to doing it by hand is identifying the main operator on the left, and undoing that operation until you get what you want. For example: (5*x)/2 = 10. First get rid of the two on the denominator by multiplying both sides by two, then get rid of multiplication by dividing by 5.

To be honest this isn't really a technique problem. It's more of a...know algebra problem.

Yeah, kind of realised, haven't had much work with algebra lately, but I'm starting to refresh my mind again.

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Write it all in one step on your calculator then.

Yeah, I would have to, more or less..

One thing is understanding what you're doing and other is doing it. That's why I wanted to kind of get the background behind it. Going to give (2) a shot.


Did (1) on the calculator now understanding what was happening and how to actually do the equation and yes, i got the right answer.

[Kezotar]

Wow, my brain hurts.

Never did get enjoy math, came here to try to understand it.

Actually this is not that complex to what many work with, however I'm just not that good at it..

Any tip on or guidance on the (2)? I can't quite get it to work with solving the equation.

- - - Updated - - -

Actually this is not that complex to what many work with, however I'm just not that good at it..

Any tip on or guidance on the (2)? I can't quite get it to work with solving the equation.

Anyone ? kinda of desperate, been trying everything but can't get around it..

[Blueobelisk]

On my TI83+ I'd put one side of the equation into y1, other side into y2, and trace the intersection of the graph.

[Kezotar]

On my TI83+ I'd put one side of the equation into y1, other side into y2, and trace the intersection of the graph.

I'm not getting it on Y1, and Y2. I try to put it in there like this;

Y1 = 30000 * (1.04^x-1)/(1.04-1)
Y2 = 400000

However, I manage to find it on Table after adjsuting the numbers, however i could be alot of work that way in the long run.. would you know what i did wrong?

[Blueobelisk]

Make sure your window is huge enough that you're going to see the intersection.

[Kezotar]

Make sure your window is huge enough that you're going to see the intersection.

WoW! I got it to work, so thanks a bunch for the help man! You really saved me out on this one!!! Btw, is this how you usually do it with this kind of equations? or would this had been possible by hand?

[Dendrek]

These are both algebra problems (the second one involves logs, so it's slightly more complicated).

Your goal is to isolate the variable (in problem 2, that's "n" obviously). You do this by trying to remove everything else that's around it using reverse order of operations. I'll use a basic example to make my point.

Let's say you had the problem:

(2^n - 1)/3 = 4

To get "n" by itself, we need to get rid of those other numbers around it (the 3 and the 1) from the outside and working our way in. The "outer operation" is the "/3". To remove that, do the opposite (of division): multiply by 3. You get 2^n - 1 = 12. Now get rid of the "- 1" by doing the opposite: add 1. You get 2^n = 13. Last, we need to get that "n" down from the exponent. Logs can do that. I'll explain the identity below:

log (N^M) => M*log(N)
In other words, when you have an exponential (N^M or 2^n are exponentials) taking the log of that will allow you to move the exponent ("M" in the first example, or "n" in the second) to the front of the log, turning the exponential problem into a multiplication problem. Going back to my example above (2^n = 13), if we take the log of both sides, we can move that "n" from the exponent.

log(2^n) = log(13) -- keep in mind that you have to take the log of BOTH sides. This keeps the equivalence statement true. (We started with 2^n = 13, which we know must be a true statement because every step that led to it was true. If that's true, then log of both sides will, most likely, also be true.) Based on the log property I posted above, we can pull that "n" down to get: n*log2 = log13. Finally, we get "n" by itself by dividing both sides by log2. So n = log13/log2.

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In your actual problem, there will be more steps, but the process will follow the same logic. If any of this doesn't make sense, please let me know.

[Kezotar]

These are both algebra problems (the second one involves logs, so it's slightly more complicated).

Your goal is to isolate the variable (in problem 2, that's "n" obviously). You do this by trying to remove everything else that's around it using reverse order of operations. I'll use a basic example to make my point.

Let's say you had the problem:

(2^n - 1)/3 = 4

To get "n" by itself, we need to get rid of those other numbers around it (the 3 and the 1) from the outside and working our way in. The "outer operation" is the "/3". To remove that, do the opposite (of division): multiply by 3. You get 2^n - 1 = 12. Now get rid of the "- 1" by doing the opposite: add 1. You get 2^n = 13. Last, we need to get that "n" down from the exponent. Logs can do that. I'll explain the identity below:

log (N^M) => M*log(N)
In other words, when you have an exponential (N^M or 2^n are exponentials) taking the log of that will allow you to move the exponent ("M" in the first example, or "n" in the second) to the front of the log, turning the exponential problem into a multiplication problem. Going back to my example above (2^n = 13), if we take the log of both sides, we can move that "n" from the exponent.

log(2^n) = log(13) -- keep in mind that you have to take the log of BOTH sides. This keeps the equivalence statement true. (We started with 2^n = 13, which we know must be a true statement because every step that led to it was true. If that's true, then log of both sides will, most likely, also be true.) Based on the log property I posted above, we can pull that "n" down to get: n*log2 = log13. Finally, we get "n" by itself by dividing both sides by log2. So n = log13/log2.

------

In your actual problem, there will be more steps, but the process will follow the same logic. If any of this doesn't make sense, please let me know.

Oh, that's alot! But yes, it kind of makes sense now, this will be handy, thanks alot man!

I even managed to solve the question 3 by the help of you two guys, thanks a alot!

[Dendrek]

No problem. Have you learned about logs? I assume you must have since you are in calculus. So hopefully everything I've explained you've seen before.

I recommend you try working it out and let me know if you get stuck. Good luck!

P.S. I realize English isn't your first language and I use a lot of technical terms. Even a native English speaker might not know what all of that means. If you need me to rewrite anything feel free to ask.