The base would be from where it previously was at. If the estimated model were:
$$ E[\log(y) \mid x] = 2 \log(x) $$
An increase in $x$ from 2 to 2.1 (an increase of 5 percent) would increase $E[y\mid x]$ from 4 to 4.41, an increase of 10.25 percent, which is pretty close to 2$\cdot$5=10 percent implied by the "log differences = percent changes" approximation.
The approximation becomes exact for infinitely small changes. The approximation is more accurate if you do small changes and update the base.
Problems with bigger changes and no base update:
For really big changes (eg 30, 40 percent or more), the approximation will start not being accurate.
For example $\log(1.4) - \log(1) \approx .33$. A 40 percent increase produces only a .33 change in the log. The approximation isn't as good. What's loosely happening is that $(1.01)^{33} \approx 1.4$, that is, 33 one percent increases adds up to a 40 percent increase.
If you need precise percent changes for some reason, you can always calculate what the percent change actually is rather than using the approximation.
The mathematics behind this approximation:
The logic of why differences in logarithms are reasonably close to percent changes can be seen through the Taylor Expansion.
The first order Taylor expansion of the logarithm function around the value of 1 is given by:
$$ \begin{align*} \log(x) &\approx \log(1) + \frac{1}{1} (x - 1) \\
&\approx x - 1 \end{align*}
$$
Hence we can write:
$$\begin{align*} \log(y_2) - \log(y_1) &= \log\left(\frac{y_2}{y_1}\right) \\
&\approx \frac{y_2 - y_1}{y_1} \quad \quad \text{for }\frac{y_2}{y_1} \text{ near } 1
\end{align*}$$
Observe that $\frac{y_2 - y_1}{y_1}$ is the percent change from $y_1$ to $y_2$. Also observe that for $y_2/y_1$ far from 1, the approximation won't be as good. For example going from $y_1=1$ to $y_2=2$ is a 100 percent increase, but $\log(2) - \log(1)$ is only .69.
