Mixing
logarithms: $a^n =b$, got $ln(x)$, but how do I get “$n$†using it?
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I learned a fast way to get $x$ in $ln(x)$ using ($n= x-1$ btw)
$$ln(x)=2left(frac12n+1+frac13(2n+1)^3+frac15(2n+1)^5+frac17(2n+1)^7+frac19(2n+1)^9...right)
$$
from wiki, which basically converts any base (i.e. $x$), like $2,$ into $e$ (base $2.71828..).$
But given something like $3^n=750$ or $2^n=1000,$ I still don't know how to get "n" by hand, and i don't understand how calculators are getting such high precision answers. $30$ decimals of precision must be coming from somewhere, and I think its math, because its too big to fit on a table, and since my computer doesn't crash or even blink when i get an answer, the number of steps can't possibly be in the millions, or likely even the thousands.
What's going on? How do I get $n,$ once I have $ln(x)$ solved?
logarithms
edited Jul 31 at 16:45
packetpacket
251112
asked Jul 31 at 15:33
Tristian
1457
add a comment |Â
up vote
-1
down vote
favorite
I learned a fast way to get $x$ in $ln(x)$ using ($n= x-1$ btw)
$$ln(x)=2left(frac12n+1+frac13(2n+1)^3+frac15(2n+1)^5+frac17(2n+1)^7+frac19(2n+1)^9...right)
$$
from wiki, which basically converts any base (i.e. $x$), like $2,$ into $e$ (base $2.71828..).$
But given something like $3^n=750$ or $2^n=1000,$ I still don't know how to get "n" by hand, and i don't understand how calculators are getting such high precision answers. $30$ decimals of precision must be coming from somewhere, and I think its math, because its too big to fit on a table, and since my computer doesn't crash or even blink when i get an answer, the number of steps can't possibly be in the millions, or likely even the thousands.
What's going on? How do I get $n,$ once I have $ln(x)$ solved?
logarithms
edited Jul 31 at 16:45
packetpacket
251112
asked Jul 31 at 15:33
Tristian
1457
1
I don't understand what you are asking. Do you want to solve $3^n=750$ for $n?$
– saulspatz
Jul 31 at 15:49
It's really not clear what you're asking. If you're asking how do computers calculate logarithms, there are lots of good methods, all numerical. For example, if you look at your series above, it converges quite rapidly; using just the terms you have shown, if $n=2$ you already get eight digits of precision.
– rogerl
Jul 31 at 15:50
By the way, don't you mean $$n=x-1over1+x?$$
– saulspatz
Jul 31 at 15:54
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
I learned a fast way to get $x$ in $ln(x)$ using ($n= x-1$ btw)
$$ln(x)=2left(frac12n+1+frac13(2n+1)^3+frac15(2n+1)^5+frac17(2n+1)^7+frac19(2n+1)^9...right)
$$
from wiki, which basically converts any base (i.e. $x$), like $2,$ into $e$ (base $2.71828..).$
But given something like $3^n=750$ or $2^n=1000,$ I still don't know how to get "n" by hand, and i don't understand how calculators are getting such high precision answers. $30$ decimals of precision must be coming from somewhere, and I think its math, because its too big to fit on a table, and since my computer doesn't crash or even blink when i get an answer, the number of steps can't possibly be in the millions, or likely even the thousands.
What's going on? How do I get $n,$ once I have $ln(x)$ solved?
logarithms
edited Jul 31 at 16:45
packetpacket
251112
asked Jul 31 at 15:33
Tristian
1457
I learned a fast way to get $x$ in $ln(x)$ using ($n= x-1$ btw)
$$ln(x)=2left(frac12n+1+frac13(2n+1)^3+frac15(2n+1)^5+frac17(2n+1)^7+frac19(2n+1)^9...right)
$$
from wiki, which basically converts any base (i.e. $x$), like $2,$ into $e$ (base $2.71828..).$
But given something like $3^n=750$ or $2^n=1000,$ I still don't know how to get "n" by hand, and i don't understand how calculators are getting such high precision answers. $30$ decimals of precision must be coming from somewhere, and I think its math, because its too big to fit on a table, and since my computer doesn't crash or even blink when i get an answer, the number of steps can't possibly be in the millions, or likely even the thousands.
What's going on? How do I get $n,$ once I have $ln(x)$ solved?
logarithms
edited Jul 31 at 16:45
packetpacket
251112
asked Jul 31 at 15:33
Tristian
1457
edited Jul 31 at 16:45
packetpacket
251112
edited Jul 31 at 16:45
packetpacket
251112
edited Jul 31 at 16:45
packetpacket
251112
251112
asked Jul 31 at 15:33
Tristian
1457
asked Jul 31 at 15:33
Tristian
1457
asked Jul 31 at 15:33
Tristian
1457
1457
1
I don't understand what you are asking. Do you want to solve $3^n=750$ for $n?$
– saulspatz
Jul 31 at 15:49
It's really not clear what you're asking. If you're asking how do computers calculate logarithms, there are lots of good methods, all numerical. For example, if you look at your series above, it converges quite rapidly; using just the terms you have shown, if $n=2$ you already get eight digits of precision.
– rogerl
Jul 31 at 15:50
By the way, don't you mean $$n=x-1over1+x?$$
– saulspatz
Jul 31 at 15:54
add a comment |Â
1
I don't understand what you are asking. Do you want to solve $3^n=750$ for $n?$
– saulspatz
Jul 31 at 15:49
It's really not clear what you're asking. If you're asking how do computers calculate logarithms, there are lots of good methods, all numerical. For example, if you look at your series above, it converges quite rapidly; using just the terms you have shown, if $n=2$ you already get eight digits of precision.
– rogerl
Jul 31 at 15:50
By the way, don't you mean $$n=x-1over1+x?$$
– saulspatz
Jul 31 at 15:54
1
1
I don't understand what you are asking. Do you want to solve $3^n=750$ for $n?$
– saulspatz
Jul 31 at 15:49
I don't understand what you are asking. Do you want to solve $3^n=750$ for $n?$
– saulspatz
Jul 31 at 15:49
It's really not clear what you're asking. If you're asking how do computers calculate logarithms, there are lots of good methods, all numerical. For example, if you look at your series above, it converges quite rapidly; using just the terms you have shown, if $n=2$ you already get eight digits of precision.
– rogerl
Jul 31 at 15:50
It's really not clear what you're asking. If you're asking how do computers calculate logarithms, there are lots of good methods, all numerical. For example, if you look at your series above, it converges quite rapidly; using just the terms you have shown, if $n=2$ you already get eight digits of precision.
– rogerl
Jul 31 at 15:50
By the way, don't you mean $$n=x-1over1+x?$$
– saulspatz
Jul 31 at 15:54
By the way, don't you mean $$n=x-1over1+x?$$
– saulspatz
Jul 31 at 15:54
add a comment |Â
1 Answer
1
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oldest
votes
up vote
0
down vote
If you want to perform a change of basis, from base $a$ to $e$. Where $a>0$.
$$log_ax = fracln(x)ln(a)$$ And using your expansion this should work. Where in your examples $a=2$ or $a=3$
answered Jul 31 at 16:57
valer
3531213
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If you want to perform a change of basis, from base $a$ to $e$. Where $a>0$.
$$log_ax = fracln(x)ln(a)$$ And using your expansion this should work. Where in your examples $a=2$ or $a=3$
answered Jul 31 at 16:57
valer
3531213
add a comment |Â
up vote
0
down vote
If you want to perform a change of basis, from base $a$ to $e$. Where $a>0$.
$$log_ax = fracln(x)ln(a)$$ And using your expansion this should work. Where in your examples $a=2$ or $a=3$
answered Jul 31 at 16:57
valer
3531213
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If you want to perform a change of basis, from base $a$ to $e$. Where $a>0$.
$$log_ax = fracln(x)ln(a)$$ And using your expansion this should work. Where in your examples $a=2$ or $a=3$
answered Jul 31 at 16:57
valer
3531213
If you want to perform a change of basis, from base $a$ to $e$. Where $a>0$.
$$log_ax = fracln(x)ln(a)$$ And using your expansion this should work. Where in your examples $a=2$ or $a=3$
answered Jul 31 at 16:57
valer
3531213
answered Jul 31 at 16:57
valer
3531213
answered Jul 31 at 16:57
valer
3531213
answered Jul 31 at 16:57
valer
3531213
3531213
add a comment |Â
add a comment |Â
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1
I don't understand what you are asking. Do you want to solve $3^n=750$ for $n?$
– saulspatz
Jul 31 at 15:49
It's really not clear what you're asking. If you're asking how do computers calculate logarithms, there are lots of good methods, all numerical. For example, if you look at your series above, it converges quite rapidly; using just the terms you have shown, if $n=2$ you already get eight digits of precision.
– rogerl
Jul 31 at 15:50
By the way, don't you mean $$n=x-1over1+x?$$
– saulspatz
Jul 31 at 15:54