Elasticity and logarithms
Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$
Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:
$$ eta = b = frac{ d log y}{d log x} $$
Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?
Here's a snapshot from the book:
elasticity
asked 1 hour ago
WorldGovWorldGov
378114
add a comment |
Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$
Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:
$$ eta = b = frac{ d log y}{d log x} $$
Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?
Here's a snapshot from the book:
elasticity
asked 1 hour ago
WorldGovWorldGov
378114
1
About half of your questions have been answered, consider accepting some.
– denesp
1 hour ago
Thank you for pointing that out. I'll do so right away.
– WorldGov
53 mins ago
add a comment |
Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$
Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:
$$ eta = b = frac{ d log y}{d log x} $$
Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?
Here's a snapshot from the book:
elasticity
asked 1 hour ago
WorldGovWorldGov
378114
Let's consider a relationship between $ y $ and $ x $, $ y = a x^b $. Taking log on both sides, we have $$ log y = log a + b log x $$
Now, my textbook, Nicholson and Snyder's Basic Principles and Extensions derives the relationship between elasticity and the logarithm of the two variables thus:
$$ eta = b = frac{ d log y}{d log x} $$
Now, I understand that $ d log y = frac 1y dy $ and $ d log x = frac 1x dx $. So I understand why we can write $ eta = frac {d log y}{d log x} $. What I don't understand is: why does $ b $, which is the power on the variable $ x $, equal $ eta $?
Here's a snapshot from the book:
elasticity
elasticity
asked 1 hour ago
WorldGovWorldGov
378114
asked 1 hour ago
WorldGovWorldGov
378114
asked 1 hour ago
WorldGovWorldGov
378114
asked 1 hour ago
WorldGovWorldGov
378114
asked 1 hour ago
WorldGovWorldGov
378114
378114
1
About half of your questions have been answered, consider accepting some.
– denesp
1 hour ago
Thank you for pointing that out. I'll do so right away.
– WorldGov
53 mins ago
add a comment |
1
About half of your questions have been answered, consider accepting some.
– denesp
1 hour ago
Thank you for pointing that out. I'll do so right away.
– WorldGov
53 mins ago
1
1
About half of your questions have been answered, consider accepting some.
– denesp
1 hour ago
About half of your questions have been answered, consider accepting some.
– denesp
1 hour ago
Thank you for pointing that out. I'll do so right away.
– WorldGov
53 mins ago
Thank you for pointing that out. I'll do so right away.
– WorldGov
53 mins ago
add a comment |
2 Answers
2
active
oldest
votes
Because $a$ is a parameter, and so
$$
eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
$$
answered 1 hour ago
denespdenesp
12.4k32247
add a comment |
Differentiate both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearrange:
$$frac{dy/y}{dx/x}=eta=b$$
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "591"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2feconomics.stackexchange.com%2fquestions%2f26401%2felasticity-and-logarithms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Because $a$ is a parameter, and so
$$
eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
$$
answered 1 hour ago
denespdenesp
12.4k32247
add a comment |
Because $a$ is a parameter, and so
$$
eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
$$
answered 1 hour ago
denespdenesp
12.4k32247
add a comment |
Because $a$ is a parameter, and so
$$
eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
$$
answered 1 hour ago
denespdenesp
12.4k32247
Because $a$ is a parameter, and so
$$
eta = frac{ d log y}{d log x} = frac{ d log a + b log x}{d log x} = 0 + b.
$$
answered 1 hour ago
denespdenesp
12.4k32247
answered 1 hour ago
denespdenesp
12.4k32247
answered 1 hour ago
denespdenesp
12.4k32247
answered 1 hour ago
denespdenesp
12.4k32247
12.4k32247
add a comment |
add a comment |
Differentiate both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearrange:
$$frac{dy/y}{dx/x}=eta=b$$
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
add a comment |
Differentiate both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearrange:
$$frac{dy/y}{dx/x}=eta=b$$
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
add a comment |
Differentiate both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearrange:
$$frac{dy/y}{dx/x}=eta=b$$
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
Differentiate both sides of the equation with respect to $x$, using the chain rule for the left hand side and noting that, since $a$ is a parameter, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearrange:
$$frac{dy/y}{dx/x}=eta=b$$
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
answered 12 mins ago
Adam BaileyAdam Bailey
3,6541026
3,6541026
add a comment |
add a comment |
Thanks for contributing an answer to Economics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
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
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2feconomics.stackexchange.com%2fquestions%2f26401%2felasticity-and-logarithms%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
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
Required, but never shown
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
Required, but never shown
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
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
About half of your questions have been answered, consider accepting some.
– denesp
1 hour ago
Thank you for pointing that out. I'll do so right away.
– WorldGov
53 mins ago