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What is the intuitive reasoning behind the “change of base†formula in logarithms? [duplicate]
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Intuition behind logarithm change of base
8 answers
The "change of base" formula in logarithms is:
I've seen and I understand each step of the proof, but somehow, when I see the formula as a whole, I fail to grasp it. Why is this true? How do I intuitively make sense of this?
logarithms
asked Jul 19 at 19:02
WorldGov
1495
marked as duplicate by littleO, Y. Forman, Math Lover, Robert Wolfe, José Carlos Santos Jul 19 at 22:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
1
down vote
favorite
This question already has an answer here:
Intuition behind logarithm change of base
8 answers
The "change of base" formula in logarithms is:
I've seen and I understand each step of the proof, but somehow, when I see the formula as a whole, I fail to grasp it. Why is this true? How do I intuitively make sense of this?
logarithms
asked Jul 19 at 19:02
WorldGov
1495
marked as duplicate by littleO, Y. Forman, Math Lover, Robert Wolfe, José Carlos Santos Jul 19 at 22:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What sort of intuition are you looking for? It's a scaling formula, in a sense, that allows you to use one set of log tables given another.
– Sean Roberson
Jul 19 at 19:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
This question already has an answer here:
Intuition behind logarithm change of base
8 answers
The "change of base" formula in logarithms is:
I've seen and I understand each step of the proof, but somehow, when I see the formula as a whole, I fail to grasp it. Why is this true? How do I intuitively make sense of this?
logarithms
asked Jul 19 at 19:02
WorldGov
1495
This question already has an answer here:
Intuition behind logarithm change of base
8 answers
The "change of base" formula in logarithms is:
I've seen and I understand each step of the proof, but somehow, when I see the formula as a whole, I fail to grasp it. Why is this true? How do I intuitively make sense of this?
This question already has an answer here:
Intuition behind logarithm change of base
8 answers
logarithms
asked Jul 19 at 19:02
WorldGov
1495
asked Jul 19 at 19:02
WorldGov
1495
asked Jul 19 at 19:02
WorldGov
1495
asked Jul 19 at 19:02
WorldGov
1495
1495
marked as duplicate by littleO, Y. Forman, Math Lover, Robert Wolfe, José Carlos Santos Jul 19 at 22:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by littleO, Y. Forman, Math Lover, Robert Wolfe, José Carlos Santos Jul 19 at 22:56
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
What sort of intuition are you looking for? It's a scaling formula, in a sense, that allows you to use one set of log tables given another.
– Sean Roberson
Jul 19 at 19:04
add a comment |Â
What sort of intuition are you looking for? It's a scaling formula, in a sense, that allows you to use one set of log tables given another.
– Sean Roberson
Jul 19 at 19:04
What sort of intuition are you looking for? It's a scaling formula, in a sense, that allows you to use one set of log tables given another.
– Sean Roberson
Jul 19 at 19:04
What sort of intuition are you looking for? It's a scaling formula, in a sense, that allows you to use one set of log tables given another.
– Sean Roberson
Jul 19 at 19:04
add a comment |Â
3 Answers
3
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up vote
2
down vote
This is not really how logarithms work, but it might give you what you want. If we pretend that $log_ax$ is simply $fracxa$, then you'll see that
$$fracxa=fracfracxbfracab$$
answered Jul 19 at 19:06
RayDansh
884214
1
That's a good way to satisfy OP's need to scratch his intuitional itch. +1.
– MPW
Jul 19 at 19:41
add a comment |Â
up vote
0
down vote
It's really just a straightforward application of the definition: multiply both sides by $log_ba$ to get $$log_axcdotlog_ba=log_bx$$
which just says that if we raise $b$ to the power we need to get $a$, and then raise that to the power we need to get $x$, we get $x$...
answered Jul 19 at 21:01
Chris Custer
5,4482622
add a comment |Â
up vote
0
down vote
I think it would help if we considered growth rates of GP (geometric progressions). If we have two GP with different growth rates, what is the relative ratio of the growth rates? For example, the GP $1, x, x^2, x^3, x^4, dots$ has a growth rate of $x$. The GP $1, x^2, x^4, x^6, dots$ has a growth rate of $x^2$. The second GP contains every other term of the first. Thus, we regard the second growth rate as twice the the growth rate of the first. This is analogous to the growth rates of arithmetic progressions. In general, we will regard a relative growth rate of $x^n$ to a growth rate of $x$ as $n$. In general, we will regard the relative growth rate of $x^n$ to a growth rate of $x^m$ as $n/m$. We can write this as $textrmgr(x^m,x^n) := n/m$ as a definition. We see that this just a characteristic property of logarithms. Thus, relative ratios of growth rates are similar to measuring quantities by units. If you change the measuring unit, there is a simple equation to express the measures relative to the two units. Regardless of measuring unit, the ratio of two quantities is the same because if one quantity has a certain ratio to another quantity it is the same no matter what common measure we use to measure them with. This goes back to Euclid's algorithm for finding the greatest common measure of two quantities. To summarize, given two growth rates $x^m$ and $x^n$, then the measure of the second with respect to the first using the common growth rate of $x$ is just $n/m$. This is also called the logarithm of $x^n$ using base $x^m$ or written as an equation:
$$, log_x^m(x^n) = fracnm = fraclog_x(x^n)log_x(x^m). $$
John Napier explicitly used the parallel relationship between arithmetic progressions and geometric progressions as his working definition of logarithms and to construct his logarithm tables. There is a great answer that covers this in MSE question 47927 "Motivation for Napier's Logartithms".
It ultimately goes back to the multiplicative version of Euclid's algorithm for finding the greatest common measure, and then using that common measure to find the relative measures of the two quantities. In the case of logarithms, the quantities to be compared are growth rates of GP. Please read my answer to MSE question 2491311 "Continued fraction evaluation" for an example.
answered Jul 19 at 20:59
Somos
11.6k1933
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
This is not really how logarithms work, but it might give you what you want. If we pretend that $log_ax$ is simply $fracxa$, then you'll see that
$$fracxa=fracfracxbfracab$$
answered Jul 19 at 19:06
RayDansh
884214
1
That's a good way to satisfy OP's need to scratch his intuitional itch. +1.
– MPW
Jul 19 at 19:41
add a comment |Â
up vote
2
down vote
This is not really how logarithms work, but it might give you what you want. If we pretend that $log_ax$ is simply $fracxa$, then you'll see that
$$fracxa=fracfracxbfracab$$
answered Jul 19 at 19:06
RayDansh
884214
1
That's a good way to satisfy OP's need to scratch his intuitional itch. +1.
– MPW
Jul 19 at 19:41
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This is not really how logarithms work, but it might give you what you want. If we pretend that $log_ax$ is simply $fracxa$, then you'll see that
$$fracxa=fracfracxbfracab$$
answered Jul 19 at 19:06
RayDansh
884214
This is not really how logarithms work, but it might give you what you want. If we pretend that $log_ax$ is simply $fracxa$, then you'll see that
$$fracxa=fracfracxbfracab$$
answered Jul 19 at 19:06
RayDansh
884214
answered Jul 19 at 19:06
RayDansh
884214
answered Jul 19 at 19:06
RayDansh
884214
answered Jul 19 at 19:06
RayDansh
884214
884214
1
That's a good way to satisfy OP's need to scratch his intuitional itch. +1.
– MPW
Jul 19 at 19:41
add a comment |Â
1
That's a good way to satisfy OP's need to scratch his intuitional itch. +1.
– MPW
Jul 19 at 19:41
1
1
That's a good way to satisfy OP's need to scratch his intuitional itch. +1.
– MPW
Jul 19 at 19:41
That's a good way to satisfy OP's need to scratch his intuitional itch. +1.
– MPW
Jul 19 at 19:41
add a comment |Â
up vote
0
down vote
It's really just a straightforward application of the definition: multiply both sides by $log_ba$ to get $$log_axcdotlog_ba=log_bx$$
which just says that if we raise $b$ to the power we need to get $a$, and then raise that to the power we need to get $x$, we get $x$...
answered Jul 19 at 21:01
Chris Custer
5,4482622
add a comment |Â
up vote
0
down vote
It's really just a straightforward application of the definition: multiply both sides by $log_ba$ to get $$log_axcdotlog_ba=log_bx$$
which just says that if we raise $b$ to the power we need to get $a$, and then raise that to the power we need to get $x$, we get $x$...
answered Jul 19 at 21:01
Chris Custer
5,4482622
add a comment |Â
up vote
0
down vote
up vote
0
down vote
It's really just a straightforward application of the definition: multiply both sides by $log_ba$ to get $$log_axcdotlog_ba=log_bx$$
which just says that if we raise $b$ to the power we need to get $a$, and then raise that to the power we need to get $x$, we get $x$...
answered Jul 19 at 21:01
Chris Custer
5,4482622
It's really just a straightforward application of the definition: multiply both sides by $log_ba$ to get $$log_axcdotlog_ba=log_bx$$
which just says that if we raise $b$ to the power we need to get $a$, and then raise that to the power we need to get $x$, we get $x$...
answered Jul 19 at 21:01
Chris Custer
5,4482622
answered Jul 19 at 21:01
Chris Custer
5,4482622
answered Jul 19 at 21:01
Chris Custer
5,4482622
answered Jul 19 at 21:01
Chris Custer
5,4482622
5,4482622
add a comment |Â
add a comment |Â
up vote
0
down vote
I think it would help if we considered growth rates of GP (geometric progressions). If we have two GP with different growth rates, what is the relative ratio of the growth rates? For example, the GP $1, x, x^2, x^3, x^4, dots$ has a growth rate of $x$. The GP $1, x^2, x^4, x^6, dots$ has a growth rate of $x^2$. The second GP contains every other term of the first. Thus, we regard the second growth rate as twice the the growth rate of the first. This is analogous to the growth rates of arithmetic progressions. In general, we will regard a relative growth rate of $x^n$ to a growth rate of $x$ as $n$. In general, we will regard the relative growth rate of $x^n$ to a growth rate of $x^m$ as $n/m$. We can write this as $textrmgr(x^m,x^n) := n/m$ as a definition. We see that this just a characteristic property of logarithms. Thus, relative ratios of growth rates are similar to measuring quantities by units. If you change the measuring unit, there is a simple equation to express the measures relative to the two units. Regardless of measuring unit, the ratio of two quantities is the same because if one quantity has a certain ratio to another quantity it is the same no matter what common measure we use to measure them with. This goes back to Euclid's algorithm for finding the greatest common measure of two quantities. To summarize, given two growth rates $x^m$ and $x^n$, then the measure of the second with respect to the first using the common growth rate of $x$ is just $n/m$. This is also called the logarithm of $x^n$ using base $x^m$ or written as an equation:
$$, log_x^m(x^n) = fracnm = fraclog_x(x^n)log_x(x^m). $$
John Napier explicitly used the parallel relationship between arithmetic progressions and geometric progressions as his working definition of logarithms and to construct his logarithm tables. There is a great answer that covers this in MSE question 47927 "Motivation for Napier's Logartithms".
It ultimately goes back to the multiplicative version of Euclid's algorithm for finding the greatest common measure, and then using that common measure to find the relative measures of the two quantities. In the case of logarithms, the quantities to be compared are growth rates of GP. Please read my answer to MSE question 2491311 "Continued fraction evaluation" for an example.
answered Jul 19 at 20:59
Somos
11.6k1933
add a comment |Â
up vote
0
down vote
I think it would help if we considered growth rates of GP (geometric progressions). If we have two GP with different growth rates, what is the relative ratio of the growth rates? For example, the GP $1, x, x^2, x^3, x^4, dots$ has a growth rate of $x$. The GP $1, x^2, x^4, x^6, dots$ has a growth rate of $x^2$. The second GP contains every other term of the first. Thus, we regard the second growth rate as twice the the growth rate of the first. This is analogous to the growth rates of arithmetic progressions. In general, we will regard a relative growth rate of $x^n$ to a growth rate of $x$ as $n$. In general, we will regard the relative growth rate of $x^n$ to a growth rate of $x^m$ as $n/m$. We can write this as $textrmgr(x^m,x^n) := n/m$ as a definition. We see that this just a characteristic property of logarithms. Thus, relative ratios of growth rates are similar to measuring quantities by units. If you change the measuring unit, there is a simple equation to express the measures relative to the two units. Regardless of measuring unit, the ratio of two quantities is the same because if one quantity has a certain ratio to another quantity it is the same no matter what common measure we use to measure them with. This goes back to Euclid's algorithm for finding the greatest common measure of two quantities. To summarize, given two growth rates $x^m$ and $x^n$, then the measure of the second with respect to the first using the common growth rate of $x$ is just $n/m$. This is also called the logarithm of $x^n$ using base $x^m$ or written as an equation:
$$, log_x^m(x^n) = fracnm = fraclog_x(x^n)log_x(x^m). $$
John Napier explicitly used the parallel relationship between arithmetic progressions and geometric progressions as his working definition of logarithms and to construct his logarithm tables. There is a great answer that covers this in MSE question 47927 "Motivation for Napier's Logartithms".
It ultimately goes back to the multiplicative version of Euclid's algorithm for finding the greatest common measure, and then using that common measure to find the relative measures of the two quantities. In the case of logarithms, the quantities to be compared are growth rates of GP. Please read my answer to MSE question 2491311 "Continued fraction evaluation" for an example.
answered Jul 19 at 20:59
Somos
11.6k1933
add a comment |Â
up vote
0
down vote
up vote
0
down vote
I think it would help if we considered growth rates of GP (geometric progressions). If we have two GP with different growth rates, what is the relative ratio of the growth rates? For example, the GP $1, x, x^2, x^3, x^4, dots$ has a growth rate of $x$. The GP $1, x^2, x^4, x^6, dots$ has a growth rate of $x^2$. The second GP contains every other term of the first. Thus, we regard the second growth rate as twice the the growth rate of the first. This is analogous to the growth rates of arithmetic progressions. In general, we will regard a relative growth rate of $x^n$ to a growth rate of $x$ as $n$. In general, we will regard the relative growth rate of $x^n$ to a growth rate of $x^m$ as $n/m$. We can write this as $textrmgr(x^m,x^n) := n/m$ as a definition. We see that this just a characteristic property of logarithms. Thus, relative ratios of growth rates are similar to measuring quantities by units. If you change the measuring unit, there is a simple equation to express the measures relative to the two units. Regardless of measuring unit, the ratio of two quantities is the same because if one quantity has a certain ratio to another quantity it is the same no matter what common measure we use to measure them with. This goes back to Euclid's algorithm for finding the greatest common measure of two quantities. To summarize, given two growth rates $x^m$ and $x^n$, then the measure of the second with respect to the first using the common growth rate of $x$ is just $n/m$. This is also called the logarithm of $x^n$ using base $x^m$ or written as an equation:
$$, log_x^m(x^n) = fracnm = fraclog_x(x^n)log_x(x^m). $$
John Napier explicitly used the parallel relationship between arithmetic progressions and geometric progressions as his working definition of logarithms and to construct his logarithm tables. There is a great answer that covers this in MSE question 47927 "Motivation for Napier's Logartithms".
It ultimately goes back to the multiplicative version of Euclid's algorithm for finding the greatest common measure, and then using that common measure to find the relative measures of the two quantities. In the case of logarithms, the quantities to be compared are growth rates of GP. Please read my answer to MSE question 2491311 "Continued fraction evaluation" for an example.
answered Jul 19 at 20:59
Somos
11.6k1933
I think it would help if we considered growth rates of GP (geometric progressions). If we have two GP with different growth rates, what is the relative ratio of the growth rates? For example, the GP $1, x, x^2, x^3, x^4, dots$ has a growth rate of $x$. The GP $1, x^2, x^4, x^6, dots$ has a growth rate of $x^2$. The second GP contains every other term of the first. Thus, we regard the second growth rate as twice the the growth rate of the first. This is analogous to the growth rates of arithmetic progressions. In general, we will regard a relative growth rate of $x^n$ to a growth rate of $x$ as $n$. In general, we will regard the relative growth rate of $x^n$ to a growth rate of $x^m$ as $n/m$. We can write this as $textrmgr(x^m,x^n) := n/m$ as a definition. We see that this just a characteristic property of logarithms. Thus, relative ratios of growth rates are similar to measuring quantities by units. If you change the measuring unit, there is a simple equation to express the measures relative to the two units. Regardless of measuring unit, the ratio of two quantities is the same because if one quantity has a certain ratio to another quantity it is the same no matter what common measure we use to measure them with. This goes back to Euclid's algorithm for finding the greatest common measure of two quantities. To summarize, given two growth rates $x^m$ and $x^n$, then the measure of the second with respect to the first using the common growth rate of $x$ is just $n/m$. This is also called the logarithm of $x^n$ using base $x^m$ or written as an equation:
$$, log_x^m(x^n) = fracnm = fraclog_x(x^n)log_x(x^m). $$
John Napier explicitly used the parallel relationship between arithmetic progressions and geometric progressions as his working definition of logarithms and to construct his logarithm tables. There is a great answer that covers this in MSE question 47927 "Motivation for Napier's Logartithms".
It ultimately goes back to the multiplicative version of Euclid's algorithm for finding the greatest common measure, and then using that common measure to find the relative measures of the two quantities. In the case of logarithms, the quantities to be compared are growth rates of GP. Please read my answer to MSE question 2491311 "Continued fraction evaluation" for an example.
answered Jul 19 at 20:59
Somos
11.6k1933
edited Jul 19 at 22:28
answered Jul 19 at 20:59
Somos
11.6k1933
answered Jul 19 at 20:59
Somos
11.6k1933
answered Jul 19 at 20:59
Somos
11.6k1933
11.6k1933
add a comment |Â
add a comment |Â
What sort of intuition are you looking for? It's a scaling formula, in a sense, that allows you to use one set of log tables given another.
– Sean Roberson
Jul 19 at 19:04