If the differences between consecutive masses are roughly constant, a linear model fits. If instead the ratios (each mass divided by the previous one) are roughly constant, an exponential model fits better.

Compute $23/27$, then $19.6/23$, then $16.7/19.6$, and so on. Are those quotients close to each other? Is that common ratio closer to 1, or clearly less than 1?

The data are recorded every 6 months. In an exponential model with base equal to the common ratio, how can you write the exponent so that it increases by 1 each time 6 months pass?

In Desmos, create a table. In the first column, enter the time values $0, 6, 12, 18, 24, 30$. In the second column, enter the corresponding masses $27, 23, 19.6, 16.7, 14.2, 12$. These points will appear on the graph.

In separate lines, type each option exactly as given (for example, y=27*(0.90)^(x/6), y=27-0.46x, etc.). Make sure you use x as the input variable in Desmos instead of $t$.

Look at how closely each graph passes through or near the plotted data points. Focus especially on the later times (around $x=18$ to $x=30$), where the differences between models become clearer. Choose the function whose curve best matches the shape and values of the data points.

Look at how the mass changes every 6 months:

• From 0 to 6 months: $27.0$ to $23.0$ (change of $-4.0$)
• From 6 to 12 months: $23.0$ to $19.6$ (change of $-3.4$)
• From 12 to 18 months: $19.6$ to $16.7$ (change of $-2.9$)
• From 18 to 24 months: $16.7$ to $14.2$ (change of $-2.5$)
• From 24 to 30 months: $14.2$ to $12.0$ (change of $-2.2$)

The differences are not constant; they are getting smaller. This suggests the mass is not decreasing by the same amount each period (so a linear model is not a good fit), but is instead decreasing by roughly the same percentage each period, which points to an exponential model.

To confirm exponential behavior, divide each mass by the previous one:

• $23.0/27.0 \approx 0.85$
• $19.6/23.0 \approx 0.85$
• $16.7/19.6 \approx 0.85$
• $14.2/16.7 \approx 0.85$
• $12.0/14.2 \approx 0.85$

These ratios are all close to the same number, about $0.85$. That means each 6‑month period, the log keeps about $85\%$ of its mass from the previous period. This is exactly the pattern an exponential decay model describes.

For exponential decay measured every 6 months, a natural model is

where:

• $a$ is the initial mass (when $t=0$),
• $r$ is the decay factor (the ratio from one 6‑month period to the next),
• $t/6$ is the number of 6‑month periods that have passed.

From the table, when $t=0$, $m(0)=27.0$, so $a=27$. From the ratios we found, $r$ is about $0.85$.

Using $a=27$ and $r\approx 0.85$, the model is

Among the choices, this corresponds to option C.