If $m_1$ is the mass at time $t$ and $m_2$ is the mass at time $t+10$, write equations for $\log_{10} m_1$ and $\log_{10} m_2$ from the model, then subtract them. How can you use $\log_{10} a - \log_{10} b = \log_{10}(a/b)$?

Once you have an equation of the form $\log_{10}(\text{factor}) = 1.2$, rewrite it in exponential form to find the factor. Then decide which answer choice is closest to that value.

In Desmos, type 10^(0.12*10) into an expression line. The value that Desmos outputs is the factor by which the mass is multiplied over each 10-day interval; compare this value to the answer choices and select the closest one.

The model is

The slope $0.12$ tells you how much $\log_{10} m$ changes for each 1-day increase in $t$. Over a 10-day interval, $t$ increases by 10, so the change in $\log_{10} m$ is

Let $m_1$ be the mass at the start of some 10-day interval and $m_2$ be the mass 10 days later. Then

• $\log_{10} m_1$ is the log at the start
• $\log_{10} m_2$ is the log 10 days later

From Step 1, we know

Use the log rule $\log_{10} a - \log_{10} b = \log_{10}\left(\dfrac{a}{b}\right)$ to rewrite this as

The ratio $\dfrac{m_2}{m_1}$ is exactly the factor by which the mass is multiplied over the 10-day interval.

From Step 2, we have

Rewrite this in exponential form:

On a calculator (allowed on the SAT for this type of question),

which is closest to 16. So the mass is expected to multiply by a factor of 16 every 10 days.