Mixing
Why does the exponent drop when “ln-ing” something?
When differentiating implicit functions, you are supposed to ln both sides of the equation and then solve for y'.
Now when you add ln to something like.. say $x^a$ how come when you add ln to it you end up with the a dropping down and the ln just added onto the x like this?
$$ a (ln(x))$$
calculus
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
asked Nov 22 '18 at 6:38
mingming
3165
add a comment |
When differentiating implicit functions, you are supposed to ln both sides of the equation and then solve for y'.
Now when you add ln to something like.. say $x^a$ how come when you add ln to it you end up with the a dropping down and the ln just added onto the x like this?
$$ a (ln(x))$$
calculus
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
asked Nov 22 '18 at 6:38
mingming
3165
3
$ln(x^a)=aln x$ is a fundamental property of logarithms. It's one of their main uses in practice. It is the same property as $(e^y)^a=e^{ay}$, only written differently.
– Arthur
Nov 22 '18 at 6:55
OH I should've known that, thanks! For some reason when I read "Now add ln" to both sides I just wondered why doesn't it mean encapsulate the whole function inside ln. Is that a weird thing to think? Does it mean that my understanding of logarithms/derivatives is wrong?
– ming
Nov 22 '18 at 7:23
No, not really. As far as I can see, you may have had some random spell of notational misunderstanding. It could conceivably be part of something bigger, but I see no reason to extrapolate from this to a fundamental lack of understanding logarithms just from one case.
– Arthur
Nov 22 '18 at 7:26
If you are at the level of differentiating implicit functions, you must know about the properties of the logarithm.
– Yves Daoust
Nov 23 '18 at 10:33
The rule $ln(x^a) = a ln(x)$ is a natural counterpart to the rule $ln(xy) = ln(x)+ln(y)$.
– GEdgar
Nov 23 '18 at 13:19
add a comment |
When differentiating implicit functions, you are supposed to ln both sides of the equation and then solve for y'.
Now when you add ln to something like.. say $x^a$ how come when you add ln to it you end up with the a dropping down and the ln just added onto the x like this?
$$ a (ln(x))$$
calculus
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
asked Nov 22 '18 at 6:38
mingming
3165
When differentiating implicit functions, you are supposed to ln both sides of the equation and then solve for y'.
Now when you add ln to something like.. say $x^a$ how come when you add ln to it you end up with the a dropping down and the ln just added onto the x like this?
$$ a (ln(x))$$
calculus
calculus
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
asked Nov 22 '18 at 6:38
mingming
3165
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
asked Nov 22 '18 at 6:38
mingming
3165
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
edited Nov 22 '18 at 7:46
David G. Stork
10k21332
10k21332
asked Nov 22 '18 at 6:38
mingming
3165
asked Nov 22 '18 at 6:38
mingming
3165
asked Nov 22 '18 at 6:38
mingming
3165
3165
3
$ln(x^a)=aln x$ is a fundamental property of logarithms. It's one of their main uses in practice. It is the same property as $(e^y)^a=e^{ay}$, only written differently.
– Arthur
Nov 22 '18 at 6:55
OH I should've known that, thanks! For some reason when I read "Now add ln" to both sides I just wondered why doesn't it mean encapsulate the whole function inside ln. Is that a weird thing to think? Does it mean that my understanding of logarithms/derivatives is wrong?
– ming
Nov 22 '18 at 7:23
No, not really. As far as I can see, you may have had some random spell of notational misunderstanding. It could conceivably be part of something bigger, but I see no reason to extrapolate from this to a fundamental lack of understanding logarithms just from one case.
– Arthur
Nov 22 '18 at 7:26
If you are at the level of differentiating implicit functions, you must know about the properties of the logarithm.
– Yves Daoust
Nov 23 '18 at 10:33
The rule $ln(x^a) = a ln(x)$ is a natural counterpart to the rule $ln(xy) = ln(x)+ln(y)$.
– GEdgar
Nov 23 '18 at 13:19
add a comment |
3
$ln(x^a)=aln x$ is a fundamental property of logarithms. It's one of their main uses in practice. It is the same property as $(e^y)^a=e^{ay}$, only written differently.
– Arthur
Nov 22 '18 at 6:55
OH I should've known that, thanks! For some reason when I read "Now add ln" to both sides I just wondered why doesn't it mean encapsulate the whole function inside ln. Is that a weird thing to think? Does it mean that my understanding of logarithms/derivatives is wrong?
– ming
Nov 22 '18 at 7:23
No, not really. As far as I can see, you may have had some random spell of notational misunderstanding. It could conceivably be part of something bigger, but I see no reason to extrapolate from this to a fundamental lack of understanding logarithms just from one case.
– Arthur
Nov 22 '18 at 7:26
If you are at the level of differentiating implicit functions, you must know about the properties of the logarithm.
– Yves Daoust
Nov 23 '18 at 10:33
The rule $ln(x^a) = a ln(x)$ is a natural counterpart to the rule $ln(xy) = ln(x)+ln(y)$.
– GEdgar
Nov 23 '18 at 13:19
3
3
$ln(x^a)=aln x$ is a fundamental property of logarithms. It's one of their main uses in practice. It is the same property as $(e^y)^a=e^{ay}$, only written differently.
– Arthur
Nov 22 '18 at 6:55
$ln(x^a)=aln x$ is a fundamental property of logarithms. It's one of their main uses in practice. It is the same property as $(e^y)^a=e^{ay}$, only written differently.
– Arthur
Nov 22 '18 at 6:55
OH I should've known that, thanks! For some reason when I read "Now add ln" to both sides I just wondered why doesn't it mean encapsulate the whole function inside ln. Is that a weird thing to think? Does it mean that my understanding of logarithms/derivatives is wrong?
– ming
Nov 22 '18 at 7:23
OH I should've known that, thanks! For some reason when I read "Now add ln" to both sides I just wondered why doesn't it mean encapsulate the whole function inside ln. Is that a weird thing to think? Does it mean that my understanding of logarithms/derivatives is wrong?
– ming
Nov 22 '18 at 7:23
No, not really. As far as I can see, you may have had some random spell of notational misunderstanding. It could conceivably be part of something bigger, but I see no reason to extrapolate from this to a fundamental lack of understanding logarithms just from one case.
– Arthur
Nov 22 '18 at 7:26
No, not really. As far as I can see, you may have had some random spell of notational misunderstanding. It could conceivably be part of something bigger, but I see no reason to extrapolate from this to a fundamental lack of understanding logarithms just from one case.
– Arthur
Nov 22 '18 at 7:26
If you are at the level of differentiating implicit functions, you must know about the properties of the logarithm.
– Yves Daoust
Nov 23 '18 at 10:33
If you are at the level of differentiating implicit functions, you must know about the properties of the logarithm.
– Yves Daoust
Nov 23 '18 at 10:33
The rule $ln(x^a) = a ln(x)$ is a natural counterpart to the rule $ln(xy) = ln(x)+ln(y)$.
– GEdgar
Nov 23 '18 at 13:19
The rule $ln(x^a) = a ln(x)$ is a natural counterpart to the rule $ln(xy) = ln(x)+ln(y)$.
– GEdgar
Nov 23 '18 at 13:19
add a comment |
1 Answer
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This does not only occur in calculus. It is a fundamental property of logarithms, coming from their very definition. Recall that
$$log_ab=ciff a^c=b.$$
Also recall the exponential property
$$ x^{a+b}=x^ax^b. $$
Now suppose $a=x^m$ and $b=x^n$. Then
$$ log_x ab=log_xleft(x^mx^nright)=log_xleft(x^{m+n}right)=m+n=log_x m+log_x n. $$
Now applying this property multiple times, we get that for positive integers $n$, and arbitrary base $a$,
$$ log_a(x^n)=underbrace{log_a x+log_a x+cdots+log_a x}_{ntext{ times}}=nlog_a x. $$
Of course, taking $a=e$ we get the identity in your question,
$$ ln(x^a)=aln(x).$$
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
add a comment |
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1 Answer
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1 Answer
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oldest
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oldest
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active
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votes
This does not only occur in calculus. It is a fundamental property of logarithms, coming from their very definition. Recall that
$$log_ab=ciff a^c=b.$$
Also recall the exponential property
$$ x^{a+b}=x^ax^b. $$
Now suppose $a=x^m$ and $b=x^n$. Then
$$ log_x ab=log_xleft(x^mx^nright)=log_xleft(x^{m+n}right)=m+n=log_x m+log_x n. $$
Now applying this property multiple times, we get that for positive integers $n$, and arbitrary base $a$,
$$ log_a(x^n)=underbrace{log_a x+log_a x+cdots+log_a x}_{ntext{ times}}=nlog_a x. $$
Of course, taking $a=e$ we get the identity in your question,
$$ ln(x^a)=aln(x).$$
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
add a comment |
This does not only occur in calculus. It is a fundamental property of logarithms, coming from their very definition. Recall that
$$log_ab=ciff a^c=b.$$
Also recall the exponential property
$$ x^{a+b}=x^ax^b. $$
Now suppose $a=x^m$ and $b=x^n$. Then
$$ log_x ab=log_xleft(x^mx^nright)=log_xleft(x^{m+n}right)=m+n=log_x m+log_x n. $$
Now applying this property multiple times, we get that for positive integers $n$, and arbitrary base $a$,
$$ log_a(x^n)=underbrace{log_a x+log_a x+cdots+log_a x}_{ntext{ times}}=nlog_a x. $$
Of course, taking $a=e$ we get the identity in your question,
$$ ln(x^a)=aln(x).$$
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
add a comment |
This does not only occur in calculus. It is a fundamental property of logarithms, coming from their very definition. Recall that
$$log_ab=ciff a^c=b.$$
Also recall the exponential property
$$ x^{a+b}=x^ax^b. $$
Now suppose $a=x^m$ and $b=x^n$. Then
$$ log_x ab=log_xleft(x^mx^nright)=log_xleft(x^{m+n}right)=m+n=log_x m+log_x n. $$
Now applying this property multiple times, we get that for positive integers $n$, and arbitrary base $a$,
$$ log_a(x^n)=underbrace{log_a x+log_a x+cdots+log_a x}_{ntext{ times}}=nlog_a x. $$
Of course, taking $a=e$ we get the identity in your question,
$$ ln(x^a)=aln(x).$$
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
This does not only occur in calculus. It is a fundamental property of logarithms, coming from their very definition. Recall that
$$log_ab=ciff a^c=b.$$
Also recall the exponential property
$$ x^{a+b}=x^ax^b. $$
Now suppose $a=x^m$ and $b=x^n$. Then
$$ log_x ab=log_xleft(x^mx^nright)=log_xleft(x^{m+n}right)=m+n=log_x m+log_x n. $$
Now applying this property multiple times, we get that for positive integers $n$, and arbitrary base $a$,
$$ log_a(x^n)=underbrace{log_a x+log_a x+cdots+log_a x}_{ntext{ times}}=nlog_a x. $$
Of course, taking $a=e$ we get the identity in your question,
$$ ln(x^a)=aln(x).$$
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
answered Nov 23 '18 at 10:30
YiFanYiFan
2,6691422
2,6691422
add a comment |
add a comment |
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$ln(x^a)=aln x$ is a fundamental property of logarithms. It's one of their main uses in practice. It is the same property as $(e^y)^a=e^{ay}$, only written differently.
– Arthur
Nov 22 '18 at 6:55
OH I should've known that, thanks! For some reason when I read "Now add ln" to both sides I just wondered why doesn't it mean encapsulate the whole function inside ln. Is that a weird thing to think? Does it mean that my understanding of logarithms/derivatives is wrong?
– ming
Nov 22 '18 at 7:23
No, not really. As far as I can see, you may have had some random spell of notational misunderstanding. It could conceivably be part of something bigger, but I see no reason to extrapolate from this to a fundamental lack of understanding logarithms just from one case.
– Arthur
Nov 22 '18 at 7:26
If you are at the level of differentiating implicit functions, you must know about the properties of the logarithm.
– Yves Daoust
Nov 23 '18 at 10:33
The rule $ln(x^a) = a ln(x)$ is a natural counterpart to the rule $ln(xy) = ln(x)+ln(y)$.
– GEdgar
Nov 23 '18 at 13:19