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Proof of algebraic equation
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I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:
$$a^log_b c = c^log_b a$$
I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.
algebra-precalculus logarithms
edited 5 hours ago
José Carlos Santos
127k17102189
asked 5 hours ago
Nat Porter
271
 |Â
show 1 more comment
up vote
4
down vote
favorite
I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:
$$a^log_b c = c^log_b a$$
I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.
algebra-precalculus logarithms
edited 5 hours ago
José Carlos Santos
127k17102189
asked 5 hours ago
Nat Porter
271
While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago
I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago
HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago
1
@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago
1
I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago
 |Â
show 1 more comment
up vote
4
down vote
favorite
up vote
4
down vote
favorite
I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:
$$a^log_b c = c^log_b a$$
I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.
algebra-precalculus logarithms
edited 5 hours ago
José Carlos Santos
127k17102189
asked 5 hours ago
Nat Porter
271
I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:
$$a^log_b c = c^log_b a$$
I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.
algebra-precalculus logarithms
algebra-precalculus logarithms
edited 5 hours ago
José Carlos Santos
127k17102189
asked 5 hours ago
Nat Porter
271
edited 5 hours ago
José Carlos Santos
127k17102189
asked 5 hours ago
Nat Porter
271
edited 5 hours ago
José Carlos Santos
127k17102189
edited 5 hours ago
José Carlos Santos
127k17102189
edited 5 hours ago
José Carlos Santos
127k17102189
127k17102189
asked 5 hours ago
Nat Porter
271
asked 5 hours ago
Nat Porter
271
asked 5 hours ago
Nat Porter
271
271
While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago
I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago
HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago
1
@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago
1
I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago
 |Â
show 1 more comment
While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago
I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago
HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago
1
@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago
1
I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago
While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago
While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago
I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago
I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago
HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago
HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago
1
1
@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago
@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago
1
1
I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago
I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
2
down vote
Two positive numbers are equal if and only if their logarithms (in the same base) are equal.
Compute the logarithm in base $b$ of both:
- $log_b(a^log_bc)=log_b clog_b a$
- $log_b(c^log_ba)=dotsb$
Done.
answered 4 hours ago
egreg
169k1283191
add a comment |Â
up vote
1
down vote
First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.
Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$
And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.
(For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)
.....
Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.
And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).
And $log_b (c^log_b a)=log_b alog_b c$
And that's that.
....
Another way of thinking about it is:
$a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.
answered 4 hours ago
fleablood
63k22679
add a comment |Â
up vote
0
down vote
$$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$
The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
This is the Change-of-Base Formula.
answered 4 hours ago
Larry
622318
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Two positive numbers are equal if and only if their logarithms (in the same base) are equal.
Compute the logarithm in base $b$ of both:
- $log_b(a^log_bc)=log_b clog_b a$
- $log_b(c^log_ba)=dotsb$
Done.
answered 4 hours ago
egreg
169k1283191
add a comment |Â
up vote
2
down vote
Two positive numbers are equal if and only if their logarithms (in the same base) are equal.
Compute the logarithm in base $b$ of both:
- $log_b(a^log_bc)=log_b clog_b a$
- $log_b(c^log_ba)=dotsb$
Done.
answered 4 hours ago
egreg
169k1283191
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Two positive numbers are equal if and only if their logarithms (in the same base) are equal.
Compute the logarithm in base $b$ of both:
- $log_b(a^log_bc)=log_b clog_b a$
- $log_b(c^log_ba)=dotsb$
Done.
answered 4 hours ago
egreg
169k1283191
Two positive numbers are equal if and only if their logarithms (in the same base) are equal.
Compute the logarithm in base $b$ of both:
- $log_b(a^log_bc)=log_b clog_b a$
- $log_b(c^log_ba)=dotsb$
Done.
answered 4 hours ago
egreg
169k1283191
answered 4 hours ago
egreg
169k1283191
answered 4 hours ago
egreg
169k1283191
answered 4 hours ago
egreg
169k1283191
169k1283191
add a comment |Â
add a comment |Â
up vote
1
down vote
First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.
Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$
And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.
(For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)
.....
Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.
And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).
And $log_b (c^log_b a)=log_b alog_b c$
And that's that.
....
Another way of thinking about it is:
$a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.
answered 4 hours ago
fleablood
63k22679
add a comment |Â
up vote
1
down vote
First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.
Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$
And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.
(For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)
.....
Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.
And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).
And $log_b (c^log_b a)=log_b alog_b c$
And that's that.
....
Another way of thinking about it is:
$a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.
answered 4 hours ago
fleablood
63k22679
add a comment |Â
up vote
1
down vote
up vote
1
down vote
First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.
Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$
And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.
(For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)
.....
Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.
And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).
And $log_b (c^log_b a)=log_b alog_b c$
And that's that.
....
Another way of thinking about it is:
$a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.
answered 4 hours ago
fleablood
63k22679
First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.
Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$
And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.
(For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)
.....
Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.
And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).
And $log_b (c^log_b a)=log_b alog_b c$
And that's that.
....
Another way of thinking about it is:
$a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.
answered 4 hours ago
fleablood
63k22679
edited 4 hours ago
answered 4 hours ago
fleablood
63k22679
answered 4 hours ago
fleablood
63k22679
answered 4 hours ago
fleablood
63k22679
63k22679
add a comment |Â
add a comment |Â
up vote
0
down vote
$$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$
The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
This is the Change-of-Base Formula.
answered 4 hours ago
Larry
622318
add a comment |Â
up vote
0
down vote
$$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$
The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
This is the Change-of-Base Formula.
answered 4 hours ago
Larry
622318
add a comment |Â
up vote
0
down vote
up vote
0
down vote
$$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$
The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
This is the Change-of-Base Formula.
answered 4 hours ago
Larry
622318
$$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$
The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
This is the Change-of-Base Formula.
answered 4 hours ago
Larry
622318
edited 4 hours ago
answered 4 hours ago
Larry
622318
answered 4 hours ago
Larry
622318
answered 4 hours ago
Larry
622318
622318
add a comment |Â
add a comment |Â
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While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago
I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago
HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago
1
@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago
1
I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago