Elasticity and logarithms
$begingroup$
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 Jan 13 at 13:09
WorldGovWorldGov
393114
$endgroup$
add a comment |
$begingroup$
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 Jan 13 at 13:09
WorldGovWorldGov
393114
$endgroup$
1
$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24
1
$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44
add a comment |
$begingroup$
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 Jan 13 at 13:09
WorldGovWorldGov
393114
$endgroup$
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 Jan 13 at 13:09
WorldGovWorldGov
393114
asked Jan 13 at 13:09
WorldGovWorldGov
393114
asked Jan 13 at 13:09
WorldGovWorldGov
393114
asked Jan 13 at 13:09
WorldGovWorldGov
393114
asked Jan 13 at 13:09
WorldGovWorldGov
393114
393114
1
$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24
1
$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44
add a comment |
1
$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24
1
$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44
1
1
$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24
$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24
1
1
$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44
$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 Jan 13 at 13:21
denespdenesp
12.5k32248
$endgroup$
add a comment |
$begingroup$
Differentiating 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, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearranging:
$$frac{dy/y}{dx/x}=eta=b$$
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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votes
$begingroup$
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 Jan 13 at 13:21
denespdenesp
12.5k32248
$endgroup$
add a comment |
$begingroup$
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 Jan 13 at 13:21
denespdenesp
12.5k32248
$endgroup$
add a comment |
$begingroup$
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 Jan 13 at 13:21
denespdenesp
12.5k32248
$endgroup$
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 Jan 13 at 13:21
denespdenesp
12.5k32248
answered Jan 13 at 13:21
denespdenesp
12.5k32248
answered Jan 13 at 13:21
denespdenesp
12.5k32248
answered Jan 13 at 13:21
denespdenesp
12.5k32248
12.5k32248
add a comment |
add a comment |
$begingroup$
Differentiating 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, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearranging:
$$frac{dy/y}{dx/x}=eta=b$$
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
$endgroup$
add a comment |
$begingroup$
Differentiating 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, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearranging:
$$frac{dy/y}{dx/x}=eta=b$$
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
$endgroup$
add a comment |
$begingroup$
Differentiating 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, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearranging:
$$frac{dy/y}{dx/x}=eta=b$$
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
$endgroup$
Differentiating 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, $da/dx=0$:
$$frac{1}{y}frac{dy}{dx}=bfrac{1}{x}$$
Rearranging:
$$frac{dy/y}{dx/x}=eta=b$$
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
edited Jan 13 at 17:57
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
answered Jan 13 at 14:25
Adam BaileyAdam Bailey
3,7041026
3,7041026
add a comment |
add a comment |
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$begingroup$
About half of your questions have been answered, consider accepting some.
$endgroup$
– denesp
Jan 13 at 13:24
1
$begingroup$
Thank you for pointing that out. I'll do so right away.
$endgroup$
– WorldGov
Jan 13 at 13:44