The half-life of $${}^{65}Zn$$ is 245 days. After x days, 75% of original activity remained. The value of x in days is ___ . (Nearest integer)
(Given: log 3 = 0.4771 and log 2 = 0.3010)

We need to find the number of days after which 75% of the original activity remains, given a half-life $$t_{1/2} = 245$$ days and remaining activity of 75% (or 3/4 of the original).

Using the radioactive decay formula $$N = N_0 e^{-\lambda t}$$ or equivalently $$\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/t_{1/2}},$$ we set $$\frac{3}{4} = \left(\frac{1}{2}\right)^{x/245}$$ where x is the number of days.

Taking logarithms of both sides yields $$\log\frac{3}{4} = \frac{x}{245}\log\frac{1}{2},$$ which can be expanded to $$\log 3 - \log 4 = -\frac{x}{245}\log 2$$ and numerically to $$0.4771 - 2(0.3010) = -\frac{x}{245}(0.3010),$$ giving $$0.4771 - 0.6020 = -\frac{x}{245}(0.3010)$$ and then $$-0.1249 = -\frac{x \times 0.3010}{245}.$$

Solving for x gives $$x = \frac{0.1249 \times 245}{0.3010} = \frac{30.6}{0.3010} = 101.7 \approx 102.$$ Therefore, x = 102 days.