The half-life period of a radioactive element is $\displaystyle 1.5 \times 10^{10}$ years. Calculate the time in which the activity of the element is reduced to $\displaystyle 75$% of its original value. [Given : $\displaystyle \log 2=0 \cdot 30, \log 3=0 \cdot 48, \log 4=0 \cdot 60$ ]

k = $\displaystyle 0.693$ 𝑡$\displaystyle 1$ k = $\displaystyle 1.5$ ×$\displaystyle 1010$ year−$\displaystyle 1$ t = $\displaystyle 2.303$ $\displaystyle 0.693$ × $\displaystyle 1.5$ × $\displaystyle 1010$ 𝑙𝑜𝑔$\displaystyle 100$ t = $\displaystyle 2.303$ $\displaystyle 0.693$ × $\displaystyle 1.5$ × $\displaystyle 1010$ [𝑙𝑜𝑔$\displaystyle 4$ −𝑙𝑜𝑔$\displaystyle 3$] t = $\displaystyle 2.303$ × $\displaystyle 1.5$ × $\displaystyle 1010$ × [$\displaystyle 0.60$ −$\displaystyle 0.48$] year t = $\displaystyle 0.598$ × $\displaystyle 1010$ years OR $\displaystyle 6$ × 109years