Kinetics of inversion of cane sugar
It seems as though your issue with the formula arises due to the fact that there is a feeling of not knowing what is considered to be factors that are internal or if they are external factors. To prove the formula, the method would be as follows.
The formula for the given reaction would be:
$$\ce{C12H22O11 + H2O -&gt;  C6H12O6 (glucose) +C6H12O6 (fructose)}$$
Drawing a modified version of a RICE table with optical rotation, and the case at t=$\infty$, we get:
\begin{array}{|l|c|} \hline
\text{R} &amp;\ce{C12H22O11   &amp;C6H12O6(glucose)  &amp;C6H12O6 (fructose)} &amp;\text{Optical Rotation} \\ \hline
\text{I} &amp; n &amp; 0&amp; 0 &amp; \alpha_{s}n+\mathrm E \\
\text{C} &amp; -x &amp;  x&amp;  x &amp; - \\
\text{E} &amp; n-x &amp;  x &amp; x &amp; \alpha_{s}(n-x)+ \alpha_{g}(x) + \alpha_{f}(x)+ \mathrm E \\
\infty &amp; 0 &amp; n &amp; n &amp; (\alpha_{g}+\alpha_{f})(n) +\mathrm E \\ \hline
\end{array}
Now, for a general first order reaction, the formula would be:
$$k=\dfrac{2.303}{t}\log \frac{n}{n-x}$$
Therefore, we need to find the value of $\dfrac{n}{n-x}$. Using the given data, we get:
$$\frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}=\frac{n}{n-x}$$
Therefore the formula given is true, even considering an external factor. You consider glucose and fructose to be internal.
$$k=\dfrac{2.303}{t}\log \frac{\alpha(\mathrm I)-\alpha(\infty)}{\alpha(\mathrm t)-\alpha(\infty)}$$