Could someone help me with math? - ComputerCraft Forums (archive)

H4X0RZ #1

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Posted 17 May 2016 - 06:05 PM

Hello CC community ^^

I know that this is off-topic but I really need the help of someone who knows a lot about math…

I'm learning for a class test right now which mainly is about exponential growth. Normally I have to determine only one variable which is no problem for me. But now I ran into this table:


t	0	1	 2	 3	 5
B(t)		 700		 1575

(we use "B(t) = B(0) * q^t" to describe exponential growth)
AFAIK B(0) and q are missing. How could I determine them to calculate the other three values in the table?

Thanks for taking your time to read (and possibly answer) this,

~H4X0RZ

Edited on 17 May 2016 - 04:11 PM

SquidDev #2

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Posted 17 May 2016 - 06:19 PM

Not sure it is the correct way but I think it works:

700 = B₀ * q
1575 = B₀ * q³

Therefore:
1575 = 700 * q²
and
q²=2.25, q = 1.5

Edited on 17 May 2016 - 04:20 PM

H4X0RZ #3

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1583 posts

Location Germany

Posted 17 May 2016 - 06:28 PM

SquidDev, on 17 May 2016 - 06:19 PM said:
Not sure it is the correct way but I think it works:

700 = B₀ * q
1575 = B₀ * q³

Therefore:
1575 = 700 * q²
and
q²=2.25, q = 1.5

Yes, This works! Thank you so much!

But how did you do the first "step" below "Therefore:"? Where is that coming from? (/me feels extremely stupid right now)

Edited on 17 May 2016 - 04:28 PM

SquidDev #4

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Posted 17 May 2016 - 06:40 PM

H4X0RZ, on 17 May 2016 - 06:28 PM said:
But how did you do the first "step" below "Therefore:"? Where is that coming from? (/me feels extremely stupid right now)

You can write 1575 =B0 * q³ as 1575 = B₀ * q * q²
However, you've got 700 = B₀ * q above so you can just substitute it in.

There is probably a fancier way with simultaneous equations and logarithms but this way is easier :)/>.

Edited on 17 May 2016 - 04:41 PM

H4X0RZ #5

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Location Germany

Posted 17 May 2016 - 07:41 PM

SquidDev, on 17 May 2016 - 06:40 PM said:
H4X0RZ, on 17 May 2016 - 06:28 PM said:
But how did you do the first "step" below "Therefore:"? Where is that coming from? (/me feels extremely stupid right now)

You can write 1575 =B0 * q³ as 1575 = B₀ * q * q²
However, you've got 700 = B₀ * q above so you can just substitute it in.

There is probably a fancier way with simultaneous equations and logarithms but this way is easier :)/>.

OMG! Thanks! <3

TheRockettek #6

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Posted 17 May 2016 - 07:50 PM

o.O Math *brain explodes*

Edited on 17 May 2016 - 05:57 PM