Random Guy
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- Dec 7, 2015
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First of all, I don't know if this is "beginning algebra" or not. I am posting the question in this sub-forum because I don't know where else to post it.
In my economics textbook, the following note is given:
2 To accomplish this transformation, we apply the rule that, if \(\displaystyle \, Z\, =\, X\, \times\, Y,\, \) then \(\displaystyle \, \hat{Z}\, =\, \hat{X}\, +\, \hat{Y},\, \) where a hat (^) indicates a variable's growth rate.
3 Mathematical Note: Define \(\displaystyle \, X\, \) as the composite factors of production:
. . . . .\(\displaystyle X\, =\, k^{\alpha}\, h^{1-\alpha}\)
Taking natural logarithms of this equation,
. . . . .\(\displaystyle \ln(X)\, =\, \alpha\, \times\, \ln(k)\, +\, (1\, -\, \alpha)\, \times\, \ln(h)\)
Differentiating with respect to time yields
. . . . .\(\displaystyle \hat{X}\, =\, \alpha \hat{k}\, +\, (1\, -\, \alpha) \hat{h}\)
This seems intuitive enough, but I am having difficulty actually "proving" it in my mind. For example, assume that Z = 24, X = 4, and Y = 6. Then, assume that BOTH X and Y were increased by 50 percent, so that X2 = 6 and Y2 = 9. Since both X and Y were increased by 50 percent, both X-hat and Y-hat would equal 0.5, which is the growth rate. However, given that X2 = 6 and Y2 = 9, Z2 = 54, which would mean that Z-hat = 1.25, because (54-24)/24. This does not make sense, because X-hat plus Y-hat equals 1.0, which does not match 1.25 (which is the "real" growth rate).
I am sure I am making some major mistake here, but I can't figure out what it is.
In my economics textbook, the following note is given:
2 To accomplish this transformation, we apply the rule that, if \(\displaystyle \, Z\, =\, X\, \times\, Y,\, \) then \(\displaystyle \, \hat{Z}\, =\, \hat{X}\, +\, \hat{Y},\, \) where a hat (^) indicates a variable's growth rate.
3 Mathematical Note: Define \(\displaystyle \, X\, \) as the composite factors of production:
. . . . .\(\displaystyle X\, =\, k^{\alpha}\, h^{1-\alpha}\)
Taking natural logarithms of this equation,
. . . . .\(\displaystyle \ln(X)\, =\, \alpha\, \times\, \ln(k)\, +\, (1\, -\, \alpha)\, \times\, \ln(h)\)
Differentiating with respect to time yields
. . . . .\(\displaystyle \hat{X}\, =\, \alpha \hat{k}\, +\, (1\, -\, \alpha) \hat{h}\)
This seems intuitive enough, but I am having difficulty actually "proving" it in my mind. For example, assume that Z = 24, X = 4, and Y = 6. Then, assume that BOTH X and Y were increased by 50 percent, so that X2 = 6 and Y2 = 9. Since both X and Y were increased by 50 percent, both X-hat and Y-hat would equal 0.5, which is the growth rate. However, given that X2 = 6 and Y2 = 9, Z2 = 54, which would mean that Z-hat = 1.25, because (54-24)/24. This does not make sense, because X-hat plus Y-hat equals 1.0, which does not match 1.25 (which is the "real" growth rate).
I am sure I am making some major mistake here, but I can't figure out what it is.
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