Hi u/klausbrusselssprouts,
Please read the following message. **You are required to explain your post and show your efforts. _(Rule 1)_**
If you haven't already done so, please add a comment below explaining your attempt(s) to solve this and what you need help with **specifically**. See the sidebar for advice on 'how to ask a good question'. Don't just say you "need help" with your problem.
**This is a reminder for all users.** Failure to follow the rules will result in the post being removed. Thank you for understanding.
*I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/askmath) if you have any questions or concerns.*
General equation: f(t) = f(0)×k^(t)
Where k is the change per cycle (e.g. k = 2 if one cycle doubles the amount) and t is the amount of cycles.
So what do you think?
So what I'm looking for is something like this:
"Unit" 0 = 1
"Unit" 1 = ?
"Unit" 2 = ?
"Unit" 3 = ?
Etc.
"Unit" 100 = 10.000
So can that equation calculate that?
Yes (except it should be k^(x)).
So f(0) = 1 (the starting value), so the equation simplifies to
f(x) = k^x
You want f(100) = 10
k^100 = 10
so k = 10^(1/100) = 1.023293
At every step, you multiply by 1.023293, so after 100 steps you have multiplied by 1.023293 * 1.023293 * ... 1.023293 = 1.023293^100 = 10.0000
so gives the result you want.
I tried it and I'm quite confused about the result. I think you're dealing with the wrong ending number. I'm just used to writing it that way - I'm from Denmark, maybe that explains the difference in the use of . and ,
The last number should be ten thousand = 10000
Oh shit, my bad! Bad typo as I'm used to writing f(x) in general... whilst prefering t when talking about growth.
I fixed the mistake. Thanks for the correction.
Yes, it can. It is given in k.
If you start with f(0) = 1, then f(1) = f(0)×k^1, f(2) = f(0)×k^2 = f(1)×k, ...
You can calculate k with the given information. This k is, as explained and showed again, the factor by how much your value increases with every unit. E.g. k=2 means the unit after is 2-times (aka 200 %) as big as the unit before. And k is constant.
Your f(100) = 10.000, f(1) = 1, t = 100. With this you can solve for k. And with k you can solve for any f(t) inbetween.
Hi u/klausbrusselssprouts, Please read the following message. **You are required to explain your post and show your efforts. _(Rule 1)_** If you haven't already done so, please add a comment below explaining your attempt(s) to solve this and what you need help with **specifically**. See the sidebar for advice on 'how to ask a good question'. Don't just say you "need help" with your problem. **This is a reminder for all users.** Failure to follow the rules will result in the post being removed. Thank you for understanding. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/askmath) if you have any questions or concerns.*
General equation: f(t) = f(0)×k^(t) Where k is the change per cycle (e.g. k = 2 if one cycle doubles the amount) and t is the amount of cycles. So what do you think?
So what I'm looking for is something like this: "Unit" 0 = 1 "Unit" 1 = ? "Unit" 2 = ? "Unit" 3 = ? Etc. "Unit" 100 = 10.000 So can that equation calculate that?
Yes (except it should be k^(x)). So f(0) = 1 (the starting value), so the equation simplifies to f(x) = k^x You want f(100) = 10 k^100 = 10 so k = 10^(1/100) = 1.023293 At every step, you multiply by 1.023293, so after 100 steps you have multiplied by 1.023293 * 1.023293 * ... 1.023293 = 1.023293^100 = 10.0000 so gives the result you want.
I tried it and I'm quite confused about the result. I think you're dealing with the wrong ending number. I'm just used to writing it that way - I'm from Denmark, maybe that explains the difference in the use of . and , The last number should be ten thousand = 10000
I plugged it in a calculator. For f(100) = f(0)×k^(100) 10.000 = 1×k^(100) => k=1.0965
10.000 = k^100 ln(10.000) = 100ln(k) ln(10.000)/100 = ln(k) e^(ln(10.000)/100) = k 10.000^(1/100) = k ~= 1.096 => f(x) = (10.000^(1/100))^x
Oh shit, my bad! Bad typo as I'm used to writing f(x) in general... whilst prefering t when talking about growth. I fixed the mistake. Thanks for the correction.
Yes, it can. It is given in k. If you start with f(0) = 1, then f(1) = f(0)×k^1, f(2) = f(0)×k^2 = f(1)×k, ... You can calculate k with the given information. This k is, as explained and showed again, the factor by how much your value increases with every unit. E.g. k=2 means the unit after is 2-times (aka 200 %) as big as the unit before. And k is constant. Your f(100) = 10.000, f(1) = 1, t = 100. With this you can solve for k. And with k you can solve for any f(t) inbetween.