The amount of a substance remaining in a solution reduces exponentially over time - Leaving Cert Mathematics - Question 4(a) - 2017

Next, we find the nth term of the geometric progression, which is given by:

$T_n = ar^{n-1}$

Where the first term, a, is 95 and r is the common ratio found previously. We set up the inequality:

$T_n < 0.01$

Substituting the values into the inequality:

$95 \left( \frac{9}{20} \right)^{n-1} < 0.01$

We can simplify this to:

$\left( \frac{9}{20} \right)^{n-1} < \frac{0.01}{95}$

Taking logarithms on both sides gives:

$(n-1) \log \left( \frac{9}{20} \right) < \log \left( \frac{0.01}{95} \right)$

Because ( \log \left( \frac{9}{20} \right) ) is negative, we reverse the inequality sign.

Now we simplify:

$(n - 1) > \frac{\log \left( \frac{0.01}{95} \right)}{\log \left( \frac{9}{20} \right)}$

Finally, solving this inequality will yield:

$n > 12.47$

Thus, rounding up, the first day on which the percentage of the substance will be less than 0.01% is Day 12.