Logarithmic function that outputs [0.25, 1] when $x$ is [1,$n$] [closed]
$begingroup$
I'm looking for a formula that produces a logarithmic curve that meets the following conditions:
Given a set of positive integers from 1 to $n$, produce a logarithmic function that outputs a $y$ value of 0.25 when $x$ = 1 and the a $y$ value of 1 when $x$ = $n$. I'd also like to control the curvature of the logarithmic term.
Note:
I don't have a maths background so may not be using the correct terms.
graphing-functions
asked Dec 29 '18 at 17:40
JoeJoe
61
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closed as off-topic by Eevee Trainer, user91500, clathratus, metamorphy, Martín Vacas Vignolo Dec 30 '18 at 12:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, user91500, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm looking for a formula that produces a logarithmic curve that meets the following conditions:
Given a set of positive integers from 1 to $n$, produce a logarithmic function that outputs a $y$ value of 0.25 when $x$ = 1 and the a $y$ value of 1 when $x$ = $n$. I'd also like to control the curvature of the logarithmic term.
Note:
I don't have a maths background so may not be using the correct terms.
graphing-functions
asked Dec 29 '18 at 17:40
JoeJoe
61
$endgroup$
closed as off-topic by Eevee Trainer, user91500, clathratus, metamorphy, Martín Vacas Vignolo Dec 30 '18 at 12:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, user91500, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm looking for a formula that produces a logarithmic curve that meets the following conditions:
Given a set of positive integers from 1 to $n$, produce a logarithmic function that outputs a $y$ value of 0.25 when $x$ = 1 and the a $y$ value of 1 when $x$ = $n$. I'd also like to control the curvature of the logarithmic term.
Note:
I don't have a maths background so may not be using the correct terms.
graphing-functions
asked Dec 29 '18 at 17:40
JoeJoe
61
$endgroup$
I'm looking for a formula that produces a logarithmic curve that meets the following conditions:
Given a set of positive integers from 1 to $n$, produce a logarithmic function that outputs a $y$ value of 0.25 when $x$ = 1 and the a $y$ value of 1 when $x$ = $n$. I'd also like to control the curvature of the logarithmic term.
Note:
I don't have a maths background so may not be using the correct terms.
graphing-functions
graphing-functions
asked Dec 29 '18 at 17:40
JoeJoe
61
asked Dec 29 '18 at 17:40
JoeJoe
61
asked Dec 29 '18 at 17:40
JoeJoe
61
asked Dec 29 '18 at 17:40
JoeJoe
61
asked Dec 29 '18 at 17:40
JoeJoe
61
61
closed as off-topic by Eevee Trainer, user91500, clathratus, metamorphy, Martín Vacas Vignolo Dec 30 '18 at 12:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, user91500, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, user91500, clathratus, metamorphy, Martín Vacas Vignolo Dec 30 '18 at 12:38
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, user91500, clathratus
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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$begingroup$
First you can try to think of a logarithmic function that outputs $[0, 0.75]$ on $[1,n]$, you just need to add the constant $+0.25$ to correct it afterwards.
With this setup we see that $ln(x) = 0$ for $x=1$ so the lower part of the interval is already right. Now we want to correct the logarithm so that $ln(n) = 0.75$. For that we can add a constant factor that would scale the logarithm like $aln(x)$. This would also make sure that we don't influence the value for when $x=1$ since this would still be $0$ whatever $a$ is.
Now we just need to solve $aln(n) = 0.75$ and get $a = 0.75/ln(n)$.
I let you put all the parts together.
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
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add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First you can try to think of a logarithmic function that outputs $[0, 0.75]$ on $[1,n]$, you just need to add the constant $+0.25$ to correct it afterwards.
With this setup we see that $ln(x) = 0$ for $x=1$ so the lower part of the interval is already right. Now we want to correct the logarithm so that $ln(n) = 0.75$. For that we can add a constant factor that would scale the logarithm like $aln(x)$. This would also make sure that we don't influence the value for when $x=1$ since this would still be $0$ whatever $a$ is.
Now we just need to solve $aln(n) = 0.75$ and get $a = 0.75/ln(n)$.
I let you put all the parts together.
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
$endgroup$
add a comment |
$begingroup$
First you can try to think of a logarithmic function that outputs $[0, 0.75]$ on $[1,n]$, you just need to add the constant $+0.25$ to correct it afterwards.
With this setup we see that $ln(x) = 0$ for $x=1$ so the lower part of the interval is already right. Now we want to correct the logarithm so that $ln(n) = 0.75$. For that we can add a constant factor that would scale the logarithm like $aln(x)$. This would also make sure that we don't influence the value for when $x=1$ since this would still be $0$ whatever $a$ is.
Now we just need to solve $aln(n) = 0.75$ and get $a = 0.75/ln(n)$.
I let you put all the parts together.
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
$endgroup$
add a comment |
$begingroup$
First you can try to think of a logarithmic function that outputs $[0, 0.75]$ on $[1,n]$, you just need to add the constant $+0.25$ to correct it afterwards.
With this setup we see that $ln(x) = 0$ for $x=1$ so the lower part of the interval is already right. Now we want to correct the logarithm so that $ln(n) = 0.75$. For that we can add a constant factor that would scale the logarithm like $aln(x)$. This would also make sure that we don't influence the value for when $x=1$ since this would still be $0$ whatever $a$ is.
Now we just need to solve $aln(n) = 0.75$ and get $a = 0.75/ln(n)$.
I let you put all the parts together.
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
$endgroup$
First you can try to think of a logarithmic function that outputs $[0, 0.75]$ on $[1,n]$, you just need to add the constant $+0.25$ to correct it afterwards.
With this setup we see that $ln(x) = 0$ for $x=1$ so the lower part of the interval is already right. Now we want to correct the logarithm so that $ln(n) = 0.75$. For that we can add a constant factor that would scale the logarithm like $aln(x)$. This would also make sure that we don't influence the value for when $x=1$ since this would still be $0$ whatever $a$ is.
Now we just need to solve $aln(n) = 0.75$ and get $a = 0.75/ln(n)$.
I let you put all the parts together.
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
answered Dec 29 '18 at 18:01
ZubzubZubzub
3,8891922
3,8891922
add a comment |
add a comment |
