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
$endgroup$
$begingroup$
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
2 hours ago
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
asked 3 hours ago
user3776749user3776749
286210
286210
$begingroup$
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
2 hours ago
add a comment |
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
add a comment |
5 Answers
5
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 |
$begingroup$
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 |
$begingroup$
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?
$endgroup$
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
2 hours ago
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
2 hours ago
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 |
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5 Answers
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active
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5 Answers
5
active
oldest
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$begingroup$
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 1 hour ago
answered 2 hours ago
Michael RozenbergMichael Rozenberg
99.5k1590189
99.5k1590189
add a comment |
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 2 hours ago
fleabloodfleablood
69.4k22685
69.4k22685
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?
$endgroup$
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?
$endgroup$
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 2 hours ago
Chris CulterChris Culter
20.4k43584
20.4k43584
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
2 hours ago
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
2 hours ago
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
2 hours ago
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
2 hours ago
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 2 hours ago
answered 2 hours ago
J. W. TannerJ. W. Tanner
56611
56611
$begingroup$
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
2 hours ago
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
2 hours ago
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
2 hours ago
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
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
2 hours ago
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
2 hours ago
$begingroup$
Thanks. I edited to correct
$endgroup$
– J. W. Tanner
2 hours ago
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 1 hour ago
heropupheropup
63k66199
63k66199
add a comment |
add a comment |
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$begingroup$
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
2 hours ago