Why is the Ratio of $ln(x)$ and $log(x)$ a constant?
$begingroup$
I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:
$$c^{log(x)}$$ and $$x^{log(c)}$$
grow at the same rate. When figuring this out I ended up with the expression:
$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$
This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?
logarithms
$endgroup$
add a comment |
$begingroup$
I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:
$$c^{log(x)}$$ and $$x^{log(c)}$$
grow at the same rate. When figuring this out I ended up with the expression:
$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$
This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?
logarithms
$endgroup$
1
$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
Jan 27 at 7:00
$begingroup$
This is just the change of base formula for logarithms that you learn in basic algebra.
$endgroup$
– Brady Gilg
Jan 29 at 23:33
add a comment |
$begingroup$
I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:
$$c^{log(x)}$$ and $$x^{log(c)}$$
grow at the same rate. When figuring this out I ended up with the expression:
$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$
This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?
logarithms
$endgroup$
I was solving some "Big-Oh" algorithm asymptotic complexity problems, when I discovered that for some constant $c$ and some variable $x$:
$$c^{log(x)}$$ and $$x^{log(c)}$$
grow at the same rate. When figuring this out I ended up with the expression:
$$frac{ln(x)}{log(x)}=frac{ln(c)}{log(c)}approx 2.3025$$
This result was surpirsing and somewhat baffling to me. This might seem like a naive question, but could someone help me understand how the ratio of $log_{10}(x)$ and $ln(x)$ ends up being a constant value?
logarithms
logarithms
edited Jan 28 at 8:52
Asaf Karagila♦
305k32435765
305k32435765
asked Jan 27 at 6:41
user3776749user3776749
332212
332212
1
$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
Jan 27 at 7:00
$begingroup$
This is just the change of base formula for logarithms that you learn in basic algebra.
$endgroup$
– Brady Gilg
Jan 29 at 23:33
add a comment |
1
$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
Jan 27 at 7:00
$begingroup$
This is just the change of base formula for logarithms that you learn in basic algebra.
$endgroup$
– Brady Gilg
Jan 29 at 23:33
1
1
$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
Jan 27 at 7:00
$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
Jan 27 at 7:00
$begingroup$
This is just the change of base formula for logarithms that you learn in basic algebra.
$endgroup$
– Brady Gilg
Jan 29 at 23:33
$begingroup$
This is just the change of base formula for logarithms that you learn in basic algebra.
$endgroup$
– Brady Gilg
Jan 29 at 23:33
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$
$endgroup$
add a comment |
$begingroup$
Some of the answers already provided get close to a full explanation, but not quite.
Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.
So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$
Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.
$endgroup$
add a comment |
$begingroup$
Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?
$endgroup$
$begingroup$
I love the effort of guiding people to the answer.
$endgroup$
– Git Gud
Jan 27 at 14:25
add a comment |
$begingroup$
If $ln x = a_x$ and $log_{10} x = b_x$ then
$e^{a_x} = x$ and $10^{b_x} = x$.
Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.
So if $10^{k_x} =e^{k_xln 10} = x$ then......
By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......
$frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.
It's just a conversion constant and shouldn't surprise us.
This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).
$endgroup$
add a comment |
$begingroup$
Take the logarithm of both expressions:
$$
log(c^{log x})=log xlog c
qquad
log(x^{log c})=log clog x
$$
So, not only $c^{log x}$ and $x^{log c}$ grow at the same rate: they're equal, whatever base of logarithms you use.
For the second part, note that $x=e^{ln x}=b^{log_bx}$ by definition. Then
$$
ln x=log_bxln b
$$
Therefore, for $xne1$,
$$
frac{ln x}{log_bx}=ln b
$$
$endgroup$
add a comment |
$begingroup$
$log_a b$ = $log_c b over log_c a$ is a general rule.
Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.
$endgroup$
$begingroup$
Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
$endgroup$
– Rhys Hughes
Jan 27 at 6:54
1
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
Jan 27 at 7:10
add a comment |
$begingroup$
Because $e^{cx}=10^{x}$ for all $x$ and $c:=ln(10)approx 2.3025$
$endgroup$
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
});
}
});
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%2f3089210%2fwhy-is-the-ratio-of-lnx-and-logx-a-constant%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$
$endgroup$
add a comment |
$begingroup$
For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$
$endgroup$
add a comment |
$begingroup$
For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$
$endgroup$
For all $xneq1$, $x>0$ we have: $$frac{ln{x}}{log{x}}=frac{ln{x}}{frac{log_ex}{log_e{10}}}=log_e10=ln10.$$
edited Jan 27 at 7:55
answered Jan 27 at 6:57
Michael RozenbergMichael Rozenberg
104k1891196
104k1891196
add a comment |
add a comment |
$begingroup$
Some of the answers already provided get close to a full explanation, but not quite.
Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.
So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$
Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.
$endgroup$
add a comment |
$begingroup$
Some of the answers already provided get close to a full explanation, but not quite.
Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.
So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$
Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.
$endgroup$
add a comment |
$begingroup$
Some of the answers already provided get close to a full explanation, but not quite.
Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.
So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$
Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.
$endgroup$
Some of the answers already provided get close to a full explanation, but not quite.
Recall that if $b > 1$ and $x > 0$, and $$log_b x = y,$$ what this means is that $b^y = x$. In other words, the base-$b$ logarithm of $x$ is an exponent $y$ such that when the base $b$ is raised to the $y^{rm th}$ power, the result is $x$. This is the definition of a (real-valued) logarithm.
So, why is it that for two bases $a$, $b$, $$frac{log_a x}{log_b x}$$ is a constant not dependent on $x$? The reason is that $log_a x$ is an exponent, say $y$, such that $a^y = x$; and $log_b x$ is an exponent, say $w$, such that $b^w = x$; then $$b^w = x = a^y.$$ And now, raising both sides to the $1/w$ power, we get $$b = (b^w)^{1/w} = (a^y)^{1/w} = a^{y/w}.$$ So the ratio $y/w$ does not depend on $x$. In fact, again using the definition of logarithm, $y/w$ is the exponent for which the base $a$ must be raised to yield $b$; that is, we explicitly have $$frac{y}{w} = log_a b,$$ and from this, we get (with one additional algebraic step) what is known as the "change-of-base" formula $$frac{log_a x}{log_a b} = log_b x.$$
Note that only the definition of $log$ was used, and the rule for exponents $(b^m)^n = b^{mn}$.
answered Jan 27 at 8:03
heropupheropup
63.8k762102
63.8k762102
add a comment |
add a comment |
$begingroup$
Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?
$endgroup$
$begingroup$
I love the effort of guiding people to the answer.
$endgroup$
– Git Gud
Jan 27 at 14:25
add a comment |
$begingroup$
Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?
$endgroup$
$begingroup$
I love the effort of guiding people to the answer.
$endgroup$
– Git Gud
Jan 27 at 14:25
add a comment |
$begingroup$
Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?
$endgroup$
Hint: $ln(10)approx 2.3025$. Given that information, your conjecture is that $ln(x)=ln(10)log_{10}(x)$. Can you see a way to prove that?
answered Jan 27 at 6:49
Chris CulterChris Culter
21.3k43887
21.3k43887
$begingroup$
I love the effort of guiding people to the answer.
$endgroup$
– Git Gud
Jan 27 at 14:25
add a comment |
$begingroup$
I love the effort of guiding people to the answer.
$endgroup$
– Git Gud
Jan 27 at 14:25
$begingroup$
I love the effort of guiding people to the answer.
$endgroup$
– Git Gud
Jan 27 at 14:25
$begingroup$
I love the effort of guiding people to the answer.
$endgroup$
– Git Gud
Jan 27 at 14:25
add a comment |
$begingroup$
If $ln x = a_x$ and $log_{10} x = b_x$ then
$e^{a_x} = x$ and $10^{b_x} = x$.
Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.
So if $10^{k_x} =e^{k_xln 10} = x$ then......
By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......
$frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.
It's just a conversion constant and shouldn't surprise us.
This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).
$endgroup$
add a comment |
$begingroup$
If $ln x = a_x$ and $log_{10} x = b_x$ then
$e^{a_x} = x$ and $10^{b_x} = x$.
Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.
So if $10^{k_x} =e^{k_xln 10} = x$ then......
By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......
$frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.
It's just a conversion constant and shouldn't surprise us.
This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).
$endgroup$
add a comment |
$begingroup$
If $ln x = a_x$ and $log_{10} x = b_x$ then
$e^{a_x} = x$ and $10^{b_x} = x$.
Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.
So if $10^{k_x} =e^{k_xln 10} = x$ then......
By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......
$frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.
It's just a conversion constant and shouldn't surprise us.
This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).
$endgroup$
If $ln x = a_x$ and $log_{10} x = b_x$ then
$e^{a_x} = x$ and $10^{b_x} = x$.
Bear in mind $10 = e^{ln 10}$ so $10^k = (e^{ln 10})^k = e^{kln 10}$.
So if $10^{k_x} =e^{k_xln 10} = x$ then......
By definition $log_{10} x = k_x$ and $ln x = k_xln 10$ and so.......
$frac {ln x}{log_{10} x} = frac {k_xln 10}{k_x} = ln 10$.
It's just a conversion constant and shouldn't surprise us.
This is the very basis of the rule $log_b x = frac {log_a x}{log_a b}$ (note if $x$ is a variable and $b$ is a constant that is exactly your observation).
answered Jan 27 at 7:17
fleabloodfleablood
71k22686
71k22686
add a comment |
add a comment |
$begingroup$
Take the logarithm of both expressions:
$$
log(c^{log x})=log xlog c
qquad
log(x^{log c})=log clog x
$$
So, not only $c^{log x}$ and $x^{log c}$ grow at the same rate: they're equal, whatever base of logarithms you use.
For the second part, note that $x=e^{ln x}=b^{log_bx}$ by definition. Then
$$
ln x=log_bxln b
$$
Therefore, for $xne1$,
$$
frac{ln x}{log_bx}=ln b
$$
$endgroup$
add a comment |
$begingroup$
Take the logarithm of both expressions:
$$
log(c^{log x})=log xlog c
qquad
log(x^{log c})=log clog x
$$
So, not only $c^{log x}$ and $x^{log c}$ grow at the same rate: they're equal, whatever base of logarithms you use.
For the second part, note that $x=e^{ln x}=b^{log_bx}$ by definition. Then
$$
ln x=log_bxln b
$$
Therefore, for $xne1$,
$$
frac{ln x}{log_bx}=ln b
$$
$endgroup$
add a comment |
$begingroup$
Take the logarithm of both expressions:
$$
log(c^{log x})=log xlog c
qquad
log(x^{log c})=log clog x
$$
So, not only $c^{log x}$ and $x^{log c}$ grow at the same rate: they're equal, whatever base of logarithms you use.
For the second part, note that $x=e^{ln x}=b^{log_bx}$ by definition. Then
$$
ln x=log_bxln b
$$
Therefore, for $xne1$,
$$
frac{ln x}{log_bx}=ln b
$$
$endgroup$
Take the logarithm of both expressions:
$$
log(c^{log x})=log xlog c
qquad
log(x^{log c})=log clog x
$$
So, not only $c^{log x}$ and $x^{log c}$ grow at the same rate: they're equal, whatever base of logarithms you use.
For the second part, note that $x=e^{ln x}=b^{log_bx}$ by definition. Then
$$
ln x=log_bxln b
$$
Therefore, for $xne1$,
$$
frac{ln x}{log_bx}=ln b
$$
answered Jan 27 at 15:38
egregegreg
182k1485204
182k1485204
add a comment |
add a comment |
$begingroup$
$log_a b$ = $log_c b over log_c a$ is a general rule.
Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.
$endgroup$
$begingroup$
Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
$endgroup$
– Rhys Hughes
Jan 27 at 6:54
1
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
Jan 27 at 7:10
add a comment |
$begingroup$
$log_a b$ = $log_c b over log_c a$ is a general rule.
Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.
$endgroup$
$begingroup$
Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
$endgroup$
– Rhys Hughes
Jan 27 at 6:54
1
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
Jan 27 at 7:10
add a comment |
$begingroup$
$log_a b$ = $log_c b over log_c a$ is a general rule.
Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.
$endgroup$
$log_a b$ = $log_c b over log_c a$ is a general rule.
Thus $log_c a = {log_c b over log_a b}$ and your expression is the case of this when $a=10, b=x, c=e$.
edited Jan 27 at 7:08
answered Jan 27 at 6:49
J. W. TannerJ. W. Tanner
2,1021117
2,1021117
$begingroup$
Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
$endgroup$
– Rhys Hughes
Jan 27 at 6:54
1
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
Jan 27 at 7:10
add a comment |
$begingroup$
Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
$endgroup$
– Rhys Hughes
Jan 27 at 6:54
1
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
Jan 27 at 7:10
$begingroup$
Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
$endgroup$
– Rhys Hughes
Jan 27 at 6:54
$begingroup$
Ummm.... by your correct rule we have $$ln(10)=frac{log_x(10)}{log_x(e)}$$ which isn't what we're trying to prove.
$endgroup$
– Rhys Hughes
Jan 27 at 6:54
1
1
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
Jan 27 at 7:10
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
Jan 27 at 7:10
add a comment |
$begingroup$
Because $e^{cx}=10^{x}$ for all $x$ and $c:=ln(10)approx 2.3025$
$endgroup$
add a comment |
$begingroup$
Because $e^{cx}=10^{x}$ for all $x$ and $c:=ln(10)approx 2.3025$
$endgroup$
add a comment |
$begingroup$
Because $e^{cx}=10^{x}$ for all $x$ and $c:=ln(10)approx 2.3025$
$endgroup$
Because $e^{cx}=10^{x}$ for all $x$ and $c:=ln(10)approx 2.3025$
edited Jan 28 at 7:12
answered Jan 28 at 5:53
BananachBananach
3,84711429
3,84711429
add a comment |
add a comment |
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%2f3089210%2fwhy-is-the-ratio-of-lnx-and-logx-a-constant%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
1
$begingroup$
Indeed the expressions you gave are equal, not just of equal growth: $$c^{log(x)}=e^{log(c)log(x)}=x^{log(c )}$$
$endgroup$
– adfriedman
Jan 27 at 7:00
$begingroup$
This is just the change of base formula for logarithms that you learn in basic algebra.
$endgroup$
– Brady Gilg
Jan 29 at 23:33