Is this formula regarding rate of change and logs exact or an approximation?

Question

Suppose that a variable grows $5\%$ each year:

$$\frac{Y_t}{Y_{t-1}} = 1.05$$

Is it exactly equivalent (or at least aproximately equivalent) to

$$y_t = y_{t-1} + 0.05$$

where $y_t = \ln(Y_t)$ ? Why is that so?

Answer 1

No, not exactly.

You have $$\ln\left(\frac{Y_t}{Y_{t-1}}\right) = \ln(1.05) \implies$$

$$\ln(Y_t)-\ln(Y_{t-1})=\ln(1.05) \implies$$

$$y_t=y_{t-1}+\ln(1.05).$$

Now use a well-know Taylor expansion, $\ln(1+x)\approx x$, that is valid for small $x$ (the smaller $x$ is, the better the approximation) to see why $\ln(1.05)=0.04879\cdots \approx 0.05$.

Answer 2

$Y_t = 1.05 Y_{t-1} \implies \ln Y_t = \ln Y_{t-1} + \ln 1.05$

For $t$ close to $0$ $\ln (1+t) = \sum\limits_{n=1}^{\infty} \dfrac{\left(-1\right)^{n+1}}{n} t^n$. As $t=0.05$ the first term of the series is also $0.05$.

So your equality is useful for quick calculations but not totally correct.