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I understand the idea that the growth rate of a variable equals the derivative of the natural log of that variable, even so, i can't figure out how to get the following equation: L(t) = ln[L(0)] + nt. I think i have a good math foundation, but then i get lost at this and i wonder if a really do... I'd appreciate any kind of help or recomendation of math books. enter image description here

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  • $\begingroup$ Maybe you should have a look at equation 1.8 and 1.9, or at least include them in your question. Also, what textbook are we looking at? $\endgroup$
    – user18214
    Feb 27 at 11:03

2 Answers 2

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Apparently equation (1.8) claims that the time derivative of the log of labor equals $n$ (some constant). So:

$$\frac{\partial \log L}{\partial t}=n \qquad \forall t$$

It is straightforward to see that if we let $c \in \mathbb{R}$ be some constant, then the form $\log L (t) = c+nt$ satisfies the above differential equation. The interpretation of $c$ here is the value of the [log of labor at time $t=0$], which equals the log of [labor at time $t=0$], i.e. $c=\log L(0)$. Let's call $L(0) \equiv L_0$ So we see that $\log L (t)=\log L_0+nt$ is a solution of the differential equation.

One loose end: is the function we found the only solution of the differential equation? The answer is yes. This can be shown with some more math.

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At any given time, $L(t) = L_0e^{nt}$ where $L_0$ is the amount of labor you start with, often normalized to 1, $n$ is the growth rate of labor, and $t$ is time. The idea here is that labor force grows exponentially. Taking log:

$log(L(t)) = log(L_0) + log(e^{nt})= log(L_0) + nt$

using: $log(e^x) = x$

Taking derivative with respect to t on both sides would now give you $n$, the growth rate of labor force.

The same logic can be applied backwards as well.

Edit: See section 1.1 Here for a formal treatment of how discrete time growth rate : $n = \frac{L_{t+1}-L_t}{L_t}$

can be expressed equivalently in continuous terms.

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  • $\begingroup$ This does not answer the question. That those two equations are equivalent is another matter. Why do they both hold? $\endgroup$
    – user18214
    Feb 27 at 11:02
  • $\begingroup$ OP asked "i can't figure out how to get the following equation". Maybe I misunderstood. $\endgroup$
    – Rumi
    Feb 27 at 11:35

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