Radioactive substances decay over time. The mass \(M\), in grams, of a particular radioactive substance \(d\) days after the beginning of an experiment is shown in the table below.

$$\begin{array}{|c|c|} \hline \text{Number of days, } d & \text{Mass}, M \text{ (grams)} \\ \hline 0 & 45.00 \\ \hline 30 & 40.57 \\ \hline 60 & 36.58 \\ \hline 90 & 32.98 \\ \hline \end{array} $$

If this relationship is modeled by the function \(M(d)=a\cdot 10^{bd} \), which of the following could be the values of \(a\) and \(b\)?

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We can first use the row \(d=0, M=45\) to find the value of \(a\):

$$ M(0)=45 $$ $$M(d)=a\cdot 10^{bd}$$ $$ 45 = a \cdot 10^{b(0)} $$ $$ 45 = a \cdot 10^0 $$ $$ 45 = a $$

Substitute this value back into the initial equation:

$$M(d)=45\cdot 10^{bd}$$

We can use another point, such as \(d=30, M=40.57\) to find the value of \(b\):

$$ M(30)=40.57 $$ $$40.57=45\cdot 10^{b(30)}$$ $$ 40.57 = 45 \cdot 10^{30b} $$ $$ \frac{40.57}{45}=10^{30b} $$ $$ 0.90=10^{30b} $$

We can test options 3 and 4 to confirm which one is the correct choice, or we can use logarithms.

Solving with logarithms:

$$ 0.90=10^{30b} $$ $$ \log{0.90}=\log{10^{30b}} $$ $$ \log{0.90}=30b $$ $$ b = \frac{\log{0.90}}{30} $$ $$ b \approx -0.0015 $$