For a first-order reaction, the relation between the reaction rate constant (k) and time (t) for a given percentage of completion (p) is:

$$\ln\left(\frac{1}{1-p}\right) = kt$$

For t₅₀ (50% completion), p = 0.5:

$$\ln\left(\frac{1}{1-0.5}\right) = k\cdot t_{50}$$

$$\ln\left(\frac{1}{0.5}\right) = k\cdot t_{50}$$

$$\ln(2) = k\cdot t_{50}$$

For t₈₇.₅ (87.5% completion), p = 0.875:

$$\ln\left(\frac{1}{1-0.875}\right) = k\cdot t_{87.5}$$

$$\ln\left(\frac{1}{0.125}\right) = k\cdot t_{87.5}$$ $$\ln(8) = k\cdot t_{87.5}$$

Now, we need to find the relationship between t₈₇.₅ and t₅₀:

$$\frac{k\cdot t_{87.5}}{k\cdot t_{50}} = \frac{\ln(8)}{\ln(2)}$$

Since the k's cancel out, we have:

$$\frac{t_{87.5}}{t_{50}} = \frac{\ln(8)}{\ln(2)}$$

Using the property of logarithms, we get:

$$\frac{t_{87.5}}{t_{50}} = \frac{\ln(2^3)}{\ln(2)}$$

$$\frac{t_{87.5}}{t_{50}} = 3$$

So, the value of x is 3.