<p>At any given time, <span class="math-container">$L(t) = L_0e^{nt}$</span> where <span class="math-container">$L_0$</span> is the amount of labor you start with, often normalized to 1, <span class="math-container">$n$</span> is the growth rate of labor, and <span class="math-container">$t$</span> is time. The idea here is that labor force grows exponentially.
Taking log:</p>
<p><span class="math-container">$log(L(t)) = log(L_0) + log(e^{nt})= log(L_0) + nt$</span></p>
<p><strong>using: <span class="math-container">$log(e^x) = x$</span></strong></p>
<p>Taking derivative with respect to t on both sides would now give you <span class="math-container">$n$</span>, the growth rate of labor force.</p>
<p>The same logic can be applied backwards as well.</p>
<p>Edit: See section 1.1 <a href="https://www.ucl.ac.uk/%7Euctpa36/C41_note_growth_accounting.pdf" rel="nofollow noreferrer">Here</a> for a formal treatment of how discrete time growth rate : <span class="math-container">$n = \frac{L_{t+1}-L_t}{L_t}$</span></p>
<p>can be expressed equivalently in continuous terms.</p>

CC BY-SA 4.0
2022-02-27T11:02:58.370Z
This does not answer the question. That those two equations are equivalent is another matter. Why do they both hold?
user18214
CC BY-SA 4.0
2022-02-27T11:35:46.670Z
OP asked "i can't figure out how to get the following equation". Maybe I misunderstood.
Rumi
CC BY-SA 4.0
2022-02-27T04:46:24.457Z
2022-02-27T11:39:25.653Z
2022-02-27T11:39:25.653Z
Rumi
<p>Apparently equation (1.8) claims that the time derivative of the log of labor equals <span class="math-container">$n$</span> (some constant). So:</p>
<p><span class="math-container">$$\frac{\partial \log L}{\partial t}=n \qquad \forall  t$$</span></p>
<p>It is straightforward to see that if we let <span class="math-container">$c \in \mathbb{R}$</span> be some constant, then the form <span class="math-container">$\log L (t) = c+nt$</span> satisfies the above differential equation. The interpretation of <span class="math-container">$c$</span> here is the value of the [log of labor at time <span class="math-container">$t=0$</span>], which equals the log of [labor at time <span class="math-container">$t=0$</span>], i.e. <span class="math-container">$c=\log L(0)$</span>. Let's call <span class="math-container">$L(0) \equiv L_0$</span> So we see that <span class="math-container">$\log L (t)=\log L_0+nt$</span> is a solution of the differential equation.</p>
<p>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.</p>

CC BY-SA 4.0
2022-02-27T11:34:24.147Z
2022-02-27T11:41:06.463Z
2022-02-27T11:41:06.463Z
user18214
<p>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.
<a href="https://i.sstatic.net/7dQwR.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7dQwR.jpg" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2022-02-27T11:03:58.380Z
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?
user18214
<macroeconomics><mathematical-economics><solow>
How do you get to the formula L(t) = ln[L(0)] + nt on the Solow Model?
Ben
# How do you get to the formula L(t) = ln[L(0)] + nt on the Solow Model?

(Asked by: username_0 on 2022-02-27)

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](https://i.sstatic.net/7dQwR.jpg)](https://i.sstatic.net/7dQwR.jpg)

**Tags:** macroeconomics, mathematical-economics, solow

> **Comments:**
> **username_1**: 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?  

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## 2 answers

### Answer by username_1 on 2022-02-27. Score: 3 | Accepted answer

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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### Answer by username_2 on 2022-02-27. Score: 5

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](https://www.ucl.ac.uk/%7Euctpa36/C41_note_growth_accounting.pdf) 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.

> **Comments:**
> **username_1**: This does not answer the question. That those two equations are equivalent is another matter. Why do they both hold?  
> **username_2**: OP asked "i can't figure out how to get the following equation". Maybe I misunderstood.  

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