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**HallsofIvy** Another way to do this is to recognize that "half" life means that the amount is multiplied by 1/2 every 3.6 days- In T days, there are T/3.6 periods of 3.6 days each which means the initial amount is multiplied by 1/2 T/3.6 times: $\displaystyle x= 20(1/2)^{T/3.6}$. You want to solve $\displaystyle 7= 20(1/2)^{T/3.6}$ or, dividing both sides by 20, $\displaystyle \frac{7}{20}= (1/2)^{T/3.6}$. To solve that take the logarithm of both sides (it doesn't matter whether you take the natural logarithm or common logarithm) $\displaystyle log(\frac{7}{20})= (T/3.6)log(1/2)$ so [tex]T= 3.6\frac{log(7/20)}{log(1/2)}= 3.6\frac{log(20)- log(7)}{log(2)}, exactly what Soroban got.

The moral is that all logarithms and all exponentials are equivalent. It doesn't matter what base you use so use whatever is simplest.