Is there a logarithm base for which the logarithm becomes an identity function?
$begingroup$
Is there a base $b$ such that:
$$log_b x = x $$
(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)
I'm fairly certain the answer is no, but I can't find a clear justification for it.
(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)
Answer
I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!
First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:
$$ b^x=x $$
In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).
An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$
$$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$
so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).
I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).
As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.
logarithms
New contributor
$endgroup$
add a comment |
$begingroup$
Is there a base $b$ such that:
$$log_b x = x $$
(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)
I'm fairly certain the answer is no, but I can't find a clear justification for it.
(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)
Answer
I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!
First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:
$$ b^x=x $$
In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).
An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$
$$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$
so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).
I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).
As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.
logarithms
New contributor
$endgroup$
$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
2 mins ago
add a comment |
$begingroup$
Is there a base $b$ such that:
$$log_b x = x $$
(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)
I'm fairly certain the answer is no, but I can't find a clear justification for it.
(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)
Answer
I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!
First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:
$$ b^x=x $$
In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).
An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$
$$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$
so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).
I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).
As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.
logarithms
New contributor
$endgroup$
Is there a base $b$ such that:
$$log_b x = x $$
(The only one that comes to mind would be the invalid case of $log_1 1 = 1 $.)
I'm fairly certain the answer is no, but I can't find a clear justification for it.
(I don't have a strong mathematical background so an answer with the intuition would be much more helpful than any complex theorem proof.)
Answer
I'm expanding here the accepted answer by FredH in the way I interpreted it, putting it in layman's terms as much as possible (according to my very basic mathematical knowledge). I'm also drawing from other answers posted, so thanks everyone who took the time!
First, as suggested, instead of talking in terms of logarithms I'll rephrase the question as an exponentiation (which seems simpler to grasp). In that case we would be looking for a base $b$ that had an exponent that would be the same as its result:
$$ b^x=x $$
In this explanation I'll invert the terms calling the exponent the input and the result of the exponentiation the output (in the logarithm it's actually the other way around since it is the inverse function, but since we're talking about an identity function the relationship between the input and output is symmetrical).
An identity function, and hence a linear function, will have its output growing at the same rate as the input (called both $x$ here). Even if there's a particular $b$ and a particular $x$ that holds the equality (which there are, as shown in the examples of other answers), when $x$ starts growing, say, one unit at a time, in the left-hand side (LHS) it would mean multiplying the already $b^x$ by another $b$
$$ b^{x+1} = x + 1 iff b^x * b = x + 1 $$
so the LHS would be expanding $b$ times (multiplication) while the output is always growing by a fixed constant of $1$ (addition).
I originally thought of the invalid logarithm base of $1$, but not even $1$ would work for every $x$, it would keep the LHS steady while the output changes. If $b$ is bigger than one, no matter how close to $1$ it is, repeatedly multiplying by it would cause the LHS to, at some point, start (and keep) growing in each iteration by more than just $1$, surpassing the pace of the output (which will always grow at same rate). Worse yet, if $b$ is smaller than one it would mean that the LHS would start getting smaller (while the output grows).
As succinctly expressed in the answers, the identity function, or any linear function for that matter, won't fit a logarithm no matter the chosen base, since it would be like trying to fit a multiplication into an addition.
logarithms
logarithms
New contributor
New contributor
edited 1 hour ago
schomatis
New contributor
asked 5 hours ago
schomatisschomatis
83
83
New contributor
New contributor
$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
2 mins ago
add a comment |
$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
2 mins ago
$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
2 mins ago
$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
2 mins ago
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
For a function to be a logarithm, it should satisfy the law of logarithms:
$log ab = log a + log b$, for $a,b gt 0$.
If it were the identity function, this would become $ab = a + b$, which clearly is not always true.
$endgroup$
add a comment |
$begingroup$
Note that
$$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.
But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
$$log_{sqrt{2}}2=2.$$
$endgroup$
add a comment |
$begingroup$
In general
$$log_b a=c$$
is the same as
$$b^c=a$$
so you can leave logs behind and focus on solutions to
$$b^x=x$$
$endgroup$
add a comment |
$begingroup$
Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.
$endgroup$
1
$begingroup$
And why is $y=log_b x$ not a straight line?
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
$endgroup$
– Vasya
4 hours ago
$begingroup$
$frac0x$ is a constant function on a useful subset of $mathbb R$.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
If $b^k = k$ for all $k$ then
$(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.
....
Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.
Likewise $log_b b = 1$ and presumably $b ne 1$
$endgroup$
add a comment |
$begingroup$
No, it can't. For any base $b$, there is some real constant $C$, s.t.
$$
log_b x = C ln x
$$
If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
schomatis is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3140581%2fis-there-a-logarithm-base-for-which-the-logarithm-becomes-an-identity-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a function to be a logarithm, it should satisfy the law of logarithms:
$log ab = log a + log b$, for $a,b gt 0$.
If it were the identity function, this would become $ab = a + b$, which clearly is not always true.
$endgroup$
add a comment |
$begingroup$
For a function to be a logarithm, it should satisfy the law of logarithms:
$log ab = log a + log b$, for $a,b gt 0$.
If it were the identity function, this would become $ab = a + b$, which clearly is not always true.
$endgroup$
add a comment |
$begingroup$
For a function to be a logarithm, it should satisfy the law of logarithms:
$log ab = log a + log b$, for $a,b gt 0$.
If it were the identity function, this would become $ab = a + b$, which clearly is not always true.
$endgroup$
For a function to be a logarithm, it should satisfy the law of logarithms:
$log ab = log a + log b$, for $a,b gt 0$.
If it were the identity function, this would become $ab = a + b$, which clearly is not always true.
answered 5 hours ago
FredHFredH
1,248612
1,248612
add a comment |
add a comment |
$begingroup$
Note that
$$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.
But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
$$log_{sqrt{2}}2=2.$$
$endgroup$
add a comment |
$begingroup$
Note that
$$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.
But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
$$log_{sqrt{2}}2=2.$$
$endgroup$
add a comment |
$begingroup$
Note that
$$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.
But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
$$log_{sqrt{2}}2=2.$$
$endgroup$
Note that
$$log_b x=xiff b^x=xiff b=sqrt[x]{x}.$$
Since $sqrt[x]{x}$ is not a constant function, the relation cannot hold for all $x$.
But it can be true for some particular $x$. For example $b=sqrt{2}$ and we have
$$log_{sqrt{2}}2=2.$$
answered 5 hours ago
Eclipse SunEclipse Sun
7,7101437
7,7101437
add a comment |
add a comment |
$begingroup$
In general
$$log_b a=c$$
is the same as
$$b^c=a$$
so you can leave logs behind and focus on solutions to
$$b^x=x$$
$endgroup$
add a comment |
$begingroup$
In general
$$log_b a=c$$
is the same as
$$b^c=a$$
so you can leave logs behind and focus on solutions to
$$b^x=x$$
$endgroup$
add a comment |
$begingroup$
In general
$$log_b a=c$$
is the same as
$$b^c=a$$
so you can leave logs behind and focus on solutions to
$$b^x=x$$
$endgroup$
In general
$$log_b a=c$$
is the same as
$$b^c=a$$
so you can leave logs behind and focus on solutions to
$$b^x=x$$
answered 5 hours ago
Martin HansenMartin Hansen
18113
18113
add a comment |
add a comment |
$begingroup$
Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.
$endgroup$
1
$begingroup$
And why is $y=log_b x$ not a straight line?
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
$endgroup$
– Vasya
4 hours ago
$begingroup$
$frac0x$ is a constant function on a useful subset of $mathbb R$.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.
$endgroup$
1
$begingroup$
And why is $y=log_b x$ not a straight line?
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
$endgroup$
– Vasya
4 hours ago
$begingroup$
$frac0x$ is a constant function on a useful subset of $mathbb R$.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.
$endgroup$
Logically, $y=x$ is a straight line, $y=log_b x$ is not so they cannot coincide for all $x$.
answered 4 hours ago
VasyaVasya
3,9551618
3,9551618
1
$begingroup$
And why is $y=log_b x$ not a straight line?
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
$endgroup$
– Vasya
4 hours ago
$begingroup$
$frac0x$ is a constant function on a useful subset of $mathbb R$.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
1
$begingroup$
And why is $y=log_b x$ not a straight line?
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
$endgroup$
– Vasya
4 hours ago
$begingroup$
$frac0x$ is a constant function on a useful subset of $mathbb R$.
$endgroup$
– Henning Makholm
4 hours ago
1
1
$begingroup$
And why is $y=log_b x$ not a straight line?
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
And why is $y=log_b x$ not a straight line?
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
@HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
$endgroup$
– Vasya
4 hours ago
$begingroup$
@HenningMakholm: because it does not have a constant slope (derivative will be $cover x$ where $c$ is a constant but $x$ is obviously not)
$endgroup$
– Vasya
4 hours ago
$begingroup$
$frac0x$ is a constant function on a useful subset of $mathbb R$.
$endgroup$
– Henning Makholm
4 hours ago
$begingroup$
$frac0x$ is a constant function on a useful subset of $mathbb R$.
$endgroup$
– Henning Makholm
4 hours ago
add a comment |
$begingroup$
If $b^k = k$ for all $k$ then
$(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.
....
Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.
Likewise $log_b b = 1$ and presumably $b ne 1$
$endgroup$
add a comment |
$begingroup$
If $b^k = k$ for all $k$ then
$(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.
....
Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.
Likewise $log_b b = 1$ and presumably $b ne 1$
$endgroup$
add a comment |
$begingroup$
If $b^k = k$ for all $k$ then
$(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.
....
Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.
Likewise $log_b b = 1$ and presumably $b ne 1$
$endgroup$
If $b^k = k$ for all $k$ then
$(b^k)^m = b^{km}= k^m=km$ for all $k$ and $m$.
....
Actually the heck with it: $log_b 1 = 0$ always and $1 ne 0$.
Likewise $log_b b = 1$ and presumably $b ne 1$
edited 3 hours ago
answered 5 hours ago
fleabloodfleablood
72k22687
72k22687
add a comment |
add a comment |
$begingroup$
No, it can't. For any base $b$, there is some real constant $C$, s.t.
$$
log_b x = C ln x
$$
If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.
$endgroup$
add a comment |
$begingroup$
No, it can't. For any base $b$, there is some real constant $C$, s.t.
$$
log_b x = C ln x
$$
If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.
$endgroup$
add a comment |
$begingroup$
No, it can't. For any base $b$, there is some real constant $C$, s.t.
$$
log_b x = C ln x
$$
If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.
$endgroup$
No, it can't. For any base $b$, there is some real constant $C$, s.t.
$$
log_b x = C ln x
$$
If it were that this logarithm is identity function, then natural logarithm would be just $Cx$, which is clearly false.
edited 3 hours ago
answered 3 hours ago
enedilenedil
1,359620
1,359620
add a comment |
add a comment |
schomatis is a new contributor. Be nice, and check out our Code of Conduct.
schomatis is a new contributor. Be nice, and check out our Code of Conduct.
schomatis is a new contributor. Be nice, and check out our Code of Conduct.
schomatis is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3140581%2fis-there-a-logarithm-base-for-which-the-logarithm-becomes-an-identity-function%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Because $,log_b(1)=0,$ for all $,b,$ and you want $,log_b(x)=x,$ for all $,x,,$then if $,x=1,$ you must have $,0=1.$
$endgroup$
– Somos
2 mins ago