How to derive the growth rate rules from the log growth equation?

Question

These are approximations, which, I guess, only apply for small growth rates $g$ ($g$ is a function that denotes the change in its argument). $$\begin{align} g(xy) &= g(x) + g(y) \\ g(x/y) &= g(x) − g(y) \\ g(x^\alpha) &= \alpha g(x) \end{align}$$

And there are also these rules

$$\dfrac{\Delta(x*y)}{xy} = \dfrac{\Delta x}x + \dfrac{\Delta y}y$$

and

$$\dfrac{\Delta(x^\alpha)}{x^\alpha} = z*\dfrac{\Delta x}x$$

I guess $g(x)$ equals $\dfrac{\Delta x}x$? So these rules are actually the same as the above ones.

However, I don't see how this can be derived from a log growth equation (which one?). I know a natural log function of a percentage change (e.g. $1.02$) approximates a percentage change that represents growth rates very close to zero (e.g. $0.02$), but ...

Thank you

babi

Answer 1

Let $ \dot x $ be the time derivative of $x(t)$. That is $$ \dot x = \frac{d}{dt} x(t) $$

We define the growth rate of $x$ as $$ g = \frac{\dot x}{ x(t)} $$

To see this "roughly" in a discrete example, let $x(t_0) = 100 $ and suppose at the next instant $x(t_1) = 110$. Then g = .1 or 10%.

Lets try to compute your formula using our definition:

\begin{align} g(xy) &= \frac{ \dot {(xy)} }{xy} \\ &= \frac{ \dot x y + x \dot y } {xy}\\ & = \frac{\dot x y}{xy} + \frac{x \dot y}{xy} \\ &= \frac{\dot x}{x} + \frac{\dot y}{y} \\ &= g(x) + g(y) \end{align} Where the second inequality came from applying the product rule to $\dot {(xy)} $

Also note that by definiton $$ g = \dot {log (x)} = \frac{\dot x}{x} $$ That is to say our growth rate is just the time derivative of the logged variable.

We can use this definition to do another of your questions

$$ g(x^{\alpha}) = \log {\dot { (x^{\alpha})}} = \frac{\alpha x^{\alpha - 1}\dot x}{x^{\alpha}} = \frac{\alpha \dot x}{x} = \alpha g(x)$$

Answer 2

Not explicitly an answer to your question, but perhaps helpful.

I took a quick look at the link. I think it's trying to explain exponential growth without using calculus. The other questions you've asked suggest that you are learning calculus now. I think you'd be better off trying to understand this economics from scratch using calculus. If $f$ is growing at a constant rate (thinking of the independent variable as time $t$) the $$ f(t) = f(0)e^{rt}; $$ the growth rate is $r$. Its units are $1/\text{time}$.

Suppose population is growing at $2\%$ per year you can use $r=0.02$ in that formula if that rate is interpreted as "continuous compounding". If the $2\%$ is the increase in one year then in the formula you'd need $$ r = \ln(1.02) = 0.0198 $$ which is (as you understand) not much less than $0.02$. Then you could write the function as
$$ f(t) = f(0)e^{0.0198t} $$ or $$ f(t) = f(0)(1.02)^t . $$ The first is classier mathematics, the second captures the idea of $2\%$ annual growth" somewhat better.