Mixing
Justifying a common proof for the logarithmic power rule.
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I need to prove the logarithmic power rule: $$log_b(x^r) = r cdot log_b(x) hspace2.3cm$$
I have seen a large number of sources citing a similar proof, which goes like this:
beginalign*
textlet m &= log_b(x), \
x &= b^m \
x^r &= (b^m)^r \
log_b(x^r) &= log_b((b^m)^r) \
&= log_b(b^mr)& text(5)\
&= mr & text(6)\
&= rm \
log_b(x^r) &= r cdot log_b(x) & textQ.E.D \
endalign*
The problem I have with this proof is that in going from steps 5 to 6, I implicitly justify them as $$hspace0.7cm log_b(b^mr) = mr cdot log_b(b) = mr cdot 1 = mr $$
In this case, it appears that the conclusion to be proved has been used, and the proof is invalid because it is simply a case of circular reasoning.
Is everybody wrong, or have I missed something?
EDIT: If the proof is in fact broken, can someone please suggest how it could be fixed?
proof-verification
asked Jul 31 at 10:59
Oscar
18410
add a comment |Â
up vote
0
down vote
favorite
I need to prove the logarithmic power rule: $$log_b(x^r) = r cdot log_b(x) hspace2.3cm$$
I have seen a large number of sources citing a similar proof, which goes like this:
beginalign*
textlet m &= log_b(x), \
x &= b^m \
x^r &= (b^m)^r \
log_b(x^r) &= log_b((b^m)^r) \
&= log_b(b^mr)& text(5)\
&= mr & text(6)\
&= rm \
log_b(x^r) &= r cdot log_b(x) & textQ.E.D \
endalign*
The problem I have with this proof is that in going from steps 5 to 6, I implicitly justify them as $$hspace0.7cm log_b(b^mr) = mr cdot log_b(b) = mr cdot 1 = mr $$
In this case, it appears that the conclusion to be proved has been used, and the proof is invalid because it is simply a case of circular reasoning.
Is everybody wrong, or have I missed something?
EDIT: If the proof is in fact broken, can someone please suggest how it could be fixed?
proof-verification
asked Jul 31 at 10:59
Oscar
18410
1
If you do not have any problem going from the first line to the second line, then you shouldn't have one going from the fifth to the sixth, it is the same argument
– caverac
Jul 31 at 11:06
You should give the domains of $r,x.$ E.g. your 'proof' is wrong for $x=-2, r=2$
– gammatester
Jul 31 at 11:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I need to prove the logarithmic power rule: $$log_b(x^r) = r cdot log_b(x) hspace2.3cm$$
I have seen a large number of sources citing a similar proof, which goes like this:
beginalign*
textlet m &= log_b(x), \
x &= b^m \
x^r &= (b^m)^r \
log_b(x^r) &= log_b((b^m)^r) \
&= log_b(b^mr)& text(5)\
&= mr & text(6)\
&= rm \
log_b(x^r) &= r cdot log_b(x) & textQ.E.D \
endalign*
The problem I have with this proof is that in going from steps 5 to 6, I implicitly justify them as $$hspace0.7cm log_b(b^mr) = mr cdot log_b(b) = mr cdot 1 = mr $$
In this case, it appears that the conclusion to be proved has been used, and the proof is invalid because it is simply a case of circular reasoning.
Is everybody wrong, or have I missed something?
EDIT: If the proof is in fact broken, can someone please suggest how it could be fixed?
proof-verification
asked Jul 31 at 10:59
Oscar
18410
I need to prove the logarithmic power rule: $$log_b(x^r) = r cdot log_b(x) hspace2.3cm$$
I have seen a large number of sources citing a similar proof, which goes like this:
beginalign*
textlet m &= log_b(x), \
x &= b^m \
x^r &= (b^m)^r \
log_b(x^r) &= log_b((b^m)^r) \
&= log_b(b^mr)& text(5)\
&= mr & text(6)\
&= rm \
log_b(x^r) &= r cdot log_b(x) & textQ.E.D \
endalign*
The problem I have with this proof is that in going from steps 5 to 6, I implicitly justify them as $$hspace0.7cm log_b(b^mr) = mr cdot log_b(b) = mr cdot 1 = mr $$
In this case, it appears that the conclusion to be proved has been used, and the proof is invalid because it is simply a case of circular reasoning.
Is everybody wrong, or have I missed something?
EDIT: If the proof is in fact broken, can someone please suggest how it could be fixed?
proof-verification
asked Jul 31 at 10:59
Oscar
18410
asked Jul 31 at 10:59
Oscar
18410
asked Jul 31 at 10:59
Oscar
18410
asked Jul 31 at 10:59
Oscar
18410
18410
1
If you do not have any problem going from the first line to the second line, then you shouldn't have one going from the fifth to the sixth, it is the same argument
– caverac
Jul 31 at 11:06
You should give the domains of $r,x.$ E.g. your 'proof' is wrong for $x=-2, r=2$
– gammatester
Jul 31 at 11:08
add a comment |Â
1
If you do not have any problem going from the first line to the second line, then you shouldn't have one going from the fifth to the sixth, it is the same argument
– caverac
Jul 31 at 11:06
You should give the domains of $r,x.$ E.g. your 'proof' is wrong for $x=-2, r=2$
– gammatester
Jul 31 at 11:08
1
1
If you do not have any problem going from the first line to the second line, then you shouldn't have one going from the fifth to the sixth, it is the same argument
– caverac
Jul 31 at 11:06
If you do not have any problem going from the first line to the second line, then you shouldn't have one going from the fifth to the sixth, it is the same argument
– caverac
Jul 31 at 11:06
You should give the domains of $r,x.$ E.g. your 'proof' is wrong for $x=-2, r=2$
– gammatester
Jul 31 at 11:08
You should give the domains of $r,x.$ E.g. your 'proof' is wrong for $x=-2, r=2$
– gammatester
Jul 31 at 11:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
I see your point but in the fifth to sixth line the difference is that we know that "$x$" is in fact "$b$" which is the same as the base.
The point is that the definition of logarithm says
$a^x=yiff x=log_ay$ ($=log_aa^x$)
It is a bit like saying $log_101000=3$ and that you don't need any fancy properties; it follows from the definition.
answered Jul 31 at 12:17
daruma
711512
This is still conceptually difficult for me. If the property to be proved is also specified by definition, then the proof must be trivial: i.e, $$log_b(x^r) = r cdot log_b(x) & textQ.E.D$$ Surely that can't be?!
– Oscar
Jul 31 at 12:40
1
No the point is that $log_bx^r$ for general $x$ is non-trivial but if we have $log_bb^r$ that is equal to $r$ by the defintion. (The important part is $a^x=yiff x=log_a y=underlinelog_aa^x)$
– daruma
Jul 31 at 13:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
I see your point but in the fifth to sixth line the difference is that we know that "$x$" is in fact "$b$" which is the same as the base.
The point is that the definition of logarithm says
$a^x=yiff x=log_ay$ ($=log_aa^x$)
It is a bit like saying $log_101000=3$ and that you don't need any fancy properties; it follows from the definition.
answered Jul 31 at 12:17
daruma
711512
This is still conceptually difficult for me. If the property to be proved is also specified by definition, then the proof must be trivial: i.e, $$log_b(x^r) = r cdot log_b(x) & textQ.E.D$$ Surely that can't be?!
– Oscar
Jul 31 at 12:40
1
No the point is that $log_bx^r$ for general $x$ is non-trivial but if we have $log_bb^r$ that is equal to $r$ by the defintion. (The important part is $a^x=yiff x=log_a y=underlinelog_aa^x)$
– daruma
Jul 31 at 13:21
add a comment |Â
up vote
0
down vote
I see your point but in the fifth to sixth line the difference is that we know that "$x$" is in fact "$b$" which is the same as the base.
The point is that the definition of logarithm says
$a^x=yiff x=log_ay$ ($=log_aa^x$)
It is a bit like saying $log_101000=3$ and that you don't need any fancy properties; it follows from the definition.
answered Jul 31 at 12:17
daruma
711512
This is still conceptually difficult for me. If the property to be proved is also specified by definition, then the proof must be trivial: i.e, $$log_b(x^r) = r cdot log_b(x) & textQ.E.D$$ Surely that can't be?!
– Oscar
Jul 31 at 12:40
1
No the point is that $log_bx^r$ for general $x$ is non-trivial but if we have $log_bb^r$ that is equal to $r$ by the defintion. (The important part is $a^x=yiff x=log_a y=underlinelog_aa^x)$
– daruma
Jul 31 at 13:21
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I see your point but in the fifth to sixth line the difference is that we know that "$x$" is in fact "$b$" which is the same as the base.
The point is that the definition of logarithm says
$a^x=yiff x=log_ay$ ($=log_aa^x$)
It is a bit like saying $log_101000=3$ and that you don't need any fancy properties; it follows from the definition.
answered Jul 31 at 12:17
daruma
711512
I see your point but in the fifth to sixth line the difference is that we know that "$x$" is in fact "$b$" which is the same as the base.
The point is that the definition of logarithm says
$a^x=yiff x=log_ay$ ($=log_aa^x$)
It is a bit like saying $log_101000=3$ and that you don't need any fancy properties; it follows from the definition.
answered Jul 31 at 12:17
daruma
711512
answered Jul 31 at 12:17
daruma
711512
answered Jul 31 at 12:17
daruma
711512
answered Jul 31 at 12:17
daruma
711512
711512
This is still conceptually difficult for me. If the property to be proved is also specified by definition, then the proof must be trivial: i.e, $$log_b(x^r) = r cdot log_b(x) & textQ.E.D$$ Surely that can't be?!
– Oscar
Jul 31 at 12:40
1
No the point is that $log_bx^r$ for general $x$ is non-trivial but if we have $log_bb^r$ that is equal to $r$ by the defintion. (The important part is $a^x=yiff x=log_a y=underlinelog_aa^x)$
– daruma
Jul 31 at 13:21
add a comment |Â
This is still conceptually difficult for me. If the property to be proved is also specified by definition, then the proof must be trivial: i.e, $$log_b(x^r) = r cdot log_b(x) & textQ.E.D$$ Surely that can't be?!
– Oscar
Jul 31 at 12:40
1
No the point is that $log_bx^r$ for general $x$ is non-trivial but if we have $log_bb^r$ that is equal to $r$ by the defintion. (The important part is $a^x=yiff x=log_a y=underlinelog_aa^x)$
– daruma
Jul 31 at 13:21
This is still conceptually difficult for me. If the property to be proved is also specified by definition, then the proof must be trivial: i.e, $$log_b(x^r) = r cdot log_b(x) & textQ.E.D$$ Surely that can't be?!
– Oscar
Jul 31 at 12:40
This is still conceptually difficult for me. If the property to be proved is also specified by definition, then the proof must be trivial: i.e, $$log_b(x^r) = r cdot log_b(x) & textQ.E.D$$ Surely that can't be?!
– Oscar
Jul 31 at 12:40
1
1
No the point is that $log_bx^r$ for general $x$ is non-trivial but if we have $log_bb^r$ that is equal to $r$ by the defintion. (The important part is $a^x=yiff x=log_a y=underlinelog_aa^x)$
– daruma
Jul 31 at 13:21
No the point is that $log_bx^r$ for general $x$ is non-trivial but if we have $log_bb^r$ that is equal to $r$ by the defintion. (The important part is $a^x=yiff x=log_a y=underlinelog_aa^x)$
– daruma
Jul 31 at 13:21
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2867928%2fjustifying-a-common-proof-for-the-logarithmic-power-rule%23new-answer', 'question_page');
);
Post as a guest
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
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
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
1
If you do not have any problem going from the first line to the second line, then you shouldn't have one going from the fifth to the sixth, it is the same argument
– caverac
Jul 31 at 11:06
You should give the domains of $r,x.$ E.g. your 'proof' is wrong for $x=-2, r=2$
– gammatester
Jul 31 at 11:08