# How do you get to the formula L(t) = ln[L(0)] + nt on the Solow Model?

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.

• 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? Feb 27 at 11:03

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.

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.

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