Mixing
What is x in terms of p?
$begingroup$
In this equation
$$frac x{ln(x)} = p$$
How can I have $x$ as a function of $p$?
logarithms inverse-function
edited Jan 14 at 13:57
mrtaurho
5,44541237
asked Jan 14 at 13:51
Matin N.Matin N.
51
$endgroup$
add a comment |
$begingroup$
In this equation
$$frac x{ln(x)} = p$$
How can I have $x$ as a function of $p$?
logarithms inverse-function
edited Jan 14 at 13:57
mrtaurho
5,44541237
asked Jan 14 at 13:51
Matin N.Matin N.
51
$endgroup$
1
$begingroup$
I don't think the function $f(x)=frac{x}{ln x}$ has an elementary inverse. That is, the function is invertible, but there is no way to "nicely" write the inverse of it.
$endgroup$
– 5xum
Jan 14 at 13:53
2
$begingroup$
Assuming $p$ is an integer, $x$ is almost the $p$th prime, if that helps.
$endgroup$
– Arthur
Jan 14 at 13:55
add a comment |
$begingroup$
In this equation
$$frac x{ln(x)} = p$$
How can I have $x$ as a function of $p$?
logarithms inverse-function
edited Jan 14 at 13:57
mrtaurho
5,44541237
asked Jan 14 at 13:51
Matin N.Matin N.
51
$endgroup$
In this equation
$$frac x{ln(x)} = p$$
How can I have $x$ as a function of $p$?
logarithms inverse-function
logarithms inverse-function
edited Jan 14 at 13:57
mrtaurho
5,44541237
asked Jan 14 at 13:51
Matin N.Matin N.
51
edited Jan 14 at 13:57
mrtaurho
5,44541237
asked Jan 14 at 13:51
Matin N.Matin N.
51
edited Jan 14 at 13:57
mrtaurho
5,44541237
edited Jan 14 at 13:57
mrtaurho
5,44541237
edited Jan 14 at 13:57
mrtaurho
5,44541237
5,44541237
asked Jan 14 at 13:51
Matin N.Matin N.
51
asked Jan 14 at 13:51
Matin N.Matin N.
51
asked Jan 14 at 13:51
Matin N.Matin N.
51
51
1
$begingroup$
I don't think the function $f(x)=frac{x}{ln x}$ has an elementary inverse. That is, the function is invertible, but there is no way to "nicely" write the inverse of it.
$endgroup$
– 5xum
Jan 14 at 13:53
2
$begingroup$
Assuming $p$ is an integer, $x$ is almost the $p$th prime, if that helps.
$endgroup$
– Arthur
Jan 14 at 13:55
add a comment |
1
$begingroup$
I don't think the function $f(x)=frac{x}{ln x}$ has an elementary inverse. That is, the function is invertible, but there is no way to "nicely" write the inverse of it.
$endgroup$
– 5xum
Jan 14 at 13:53
2
$begingroup$
Assuming $p$ is an integer, $x$ is almost the $p$th prime, if that helps.
$endgroup$
– Arthur
Jan 14 at 13:55
1
1
$begingroup$
I don't think the function $f(x)=frac{x}{ln x}$ has an elementary inverse. That is, the function is invertible, but there is no way to "nicely" write the inverse of it.
$endgroup$
– 5xum
Jan 14 at 13:53
$begingroup$
I don't think the function $f(x)=frac{x}{ln x}$ has an elementary inverse. That is, the function is invertible, but there is no way to "nicely" write the inverse of it.
$endgroup$
– 5xum
Jan 14 at 13:53
2
2
$begingroup$
Assuming $p$ is an integer, $x$ is almost the $p$th prime, if that helps.
$endgroup$
– Arthur
Jan 14 at 13:55
$begingroup$
Assuming $p$ is an integer, $x$ is almost the $p$th prime, if that helps.
$endgroup$
– Arthur
Jan 14 at 13:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is a typical example of a transcendental equation with a solution that cannot be expressed with most widely known analytical functions (trigonometric, exponential and logarithmic).
Rule of thumb: if $x$ is found both inside and outside log/exp/sin, the equation cannot be inverted in terms of these functions.
However, the equation is common enough to have a so called "special function" which can be used to invert it: Lambert's function $x=W(y)$, which solves equation $xe^{x}=y$.
Of course, as Lambert's function doesn't usually have a button on a calculator, that does not help much, but does mean people have studied this function and have efficient means of calculating its value.
You just have to transform it into the form to apply the rule for Lambert's function;
$$ln x = frac{x}{p}$$
$$x=e^{x/p}$$
$$xe^{-x/p}=1$$
new variable $u=-x/p$
$$-pu e^{u}=1$$
$$ue^u=-1/p$$
$$u=W(-1/p)$$
$$x=-pu=-p W(-1/p)$$
Of course you have to be careful because for negative arguments, $W$ can have two branches (equation has two solutions), or no real solution at all, if the value is too negative (below $-1/e$).
answered Jan 14 at 13:56
orionorion
13.6k11837
$endgroup$
add a comment |
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1 Answer
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$begingroup$
This is a typical example of a transcendental equation with a solution that cannot be expressed with most widely known analytical functions (trigonometric, exponential and logarithmic).
Rule of thumb: if $x$ is found both inside and outside log/exp/sin, the equation cannot be inverted in terms of these functions.
However, the equation is common enough to have a so called "special function" which can be used to invert it: Lambert's function $x=W(y)$, which solves equation $xe^{x}=y$.
Of course, as Lambert's function doesn't usually have a button on a calculator, that does not help much, but does mean people have studied this function and have efficient means of calculating its value.
You just have to transform it into the form to apply the rule for Lambert's function;
$$ln x = frac{x}{p}$$
$$x=e^{x/p}$$
$$xe^{-x/p}=1$$
new variable $u=-x/p$
$$-pu e^{u}=1$$
$$ue^u=-1/p$$
$$u=W(-1/p)$$
$$x=-pu=-p W(-1/p)$$
Of course you have to be careful because for negative arguments, $W$ can have two branches (equation has two solutions), or no real solution at all, if the value is too negative (below $-1/e$).
answered Jan 14 at 13:56
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
This is a typical example of a transcendental equation with a solution that cannot be expressed with most widely known analytical functions (trigonometric, exponential and logarithmic).
Rule of thumb: if $x$ is found both inside and outside log/exp/sin, the equation cannot be inverted in terms of these functions.
However, the equation is common enough to have a so called "special function" which can be used to invert it: Lambert's function $x=W(y)$, which solves equation $xe^{x}=y$.
Of course, as Lambert's function doesn't usually have a button on a calculator, that does not help much, but does mean people have studied this function and have efficient means of calculating its value.
You just have to transform it into the form to apply the rule for Lambert's function;
$$ln x = frac{x}{p}$$
$$x=e^{x/p}$$
$$xe^{-x/p}=1$$
new variable $u=-x/p$
$$-pu e^{u}=1$$
$$ue^u=-1/p$$
$$u=W(-1/p)$$
$$x=-pu=-p W(-1/p)$$
Of course you have to be careful because for negative arguments, $W$ can have two branches (equation has two solutions), or no real solution at all, if the value is too negative (below $-1/e$).
answered Jan 14 at 13:56
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
This is a typical example of a transcendental equation with a solution that cannot be expressed with most widely known analytical functions (trigonometric, exponential and logarithmic).
Rule of thumb: if $x$ is found both inside and outside log/exp/sin, the equation cannot be inverted in terms of these functions.
However, the equation is common enough to have a so called "special function" which can be used to invert it: Lambert's function $x=W(y)$, which solves equation $xe^{x}=y$.
Of course, as Lambert's function doesn't usually have a button on a calculator, that does not help much, but does mean people have studied this function and have efficient means of calculating its value.
You just have to transform it into the form to apply the rule for Lambert's function;
$$ln x = frac{x}{p}$$
$$x=e^{x/p}$$
$$xe^{-x/p}=1$$
new variable $u=-x/p$
$$-pu e^{u}=1$$
$$ue^u=-1/p$$
$$u=W(-1/p)$$
$$x=-pu=-p W(-1/p)$$
Of course you have to be careful because for negative arguments, $W$ can have two branches (equation has two solutions), or no real solution at all, if the value is too negative (below $-1/e$).
answered Jan 14 at 13:56
orionorion
13.6k11837
$endgroup$
This is a typical example of a transcendental equation with a solution that cannot be expressed with most widely known analytical functions (trigonometric, exponential and logarithmic).
Rule of thumb: if $x$ is found both inside and outside log/exp/sin, the equation cannot be inverted in terms of these functions.
However, the equation is common enough to have a so called "special function" which can be used to invert it: Lambert's function $x=W(y)$, which solves equation $xe^{x}=y$.
Of course, as Lambert's function doesn't usually have a button on a calculator, that does not help much, but does mean people have studied this function and have efficient means of calculating its value.
You just have to transform it into the form to apply the rule for Lambert's function;
$$ln x = frac{x}{p}$$
$$x=e^{x/p}$$
$$xe^{-x/p}=1$$
new variable $u=-x/p$
$$-pu e^{u}=1$$
$$ue^u=-1/p$$
$$u=W(-1/p)$$
$$x=-pu=-p W(-1/p)$$
Of course you have to be careful because for negative arguments, $W$ can have two branches (equation has two solutions), or no real solution at all, if the value is too negative (below $-1/e$).
answered Jan 14 at 13:56
orionorion
13.6k11837
edited Jan 14 at 14:52
answered Jan 14 at 13:56
orionorion
13.6k11837
answered Jan 14 at 13:56
orionorion
13.6k11837
answered Jan 14 at 13:56
orionorion
13.6k11837
13.6k11837
add a comment |
add a comment |
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$begingroup$
I don't think the function $f(x)=frac{x}{ln x}$ has an elementary inverse. That is, the function is invertible, but there is no way to "nicely" write the inverse of it.
$endgroup$
– 5xum
Jan 14 at 13:53
2
$begingroup$
Assuming $p$ is an integer, $x$ is almost the $p$th prime, if that helps.
$endgroup$
– Arthur
Jan 14 at 13:55