Why does the exponent drop when “ln-ing” something?
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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 at 7:46
David G. Stork
9,21221232
asked Nov 22 at 6:38
ming
634
add a comment |
up vote
0
down vote
favorite
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 at 7:46
David G. Stork
9,21221232
asked Nov 22 at 6:38
ming
634
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 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 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 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 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 at 13:19
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 at 7:46
David G. Stork
9,21221232
asked Nov 22 at 6:38
ming
634
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 at 7:46
David G. Stork
9,21221232
asked Nov 22 at 6:38
ming
634
edited Nov 22 at 7:46
David G. Stork
9,21221232
asked Nov 22 at 6:38
ming
634
edited Nov 22 at 7:46
David G. Stork
9,21221232
edited Nov 22 at 7:46
David G. Stork
9,21221232
edited Nov 22 at 7:46
David G. Stork
9,21221232
9,21221232
asked Nov 22 at 6:38
ming
634
asked Nov 22 at 6:38
ming
634
asked Nov 22 at 6:38
ming
634
634
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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 at 13:19
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
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 at 10:30
YiFan
1,6181314
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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 at 10:30
YiFan
1,6181314
add a comment |
up vote
0
down vote
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 at 10:30
YiFan
1,6181314
add a comment |
up vote
0
down vote
up vote
0
down vote
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 at 10:30
YiFan
1,6181314
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 at 10:30
YiFan
1,6181314
answered Nov 23 at 10:30
YiFan
1,6181314
answered Nov 23 at 10:30
YiFan
1,6181314
answered Nov 23 at 10:30
YiFan
1,6181314
1,6181314
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 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 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 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 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 at 13:19