If you know $\ln(m)$ equals some expression, what operation can you apply to both sides to isolate $m$? Recall that $\ln$ is a logarithm with base $e$.

Take your equation $\ln(m) = 4.6 - 0.029t$ and apply the same operation to both sides. What does $e^{\ln(m)}$ simplify to?

Enter the line from the problem as y = 4.6 - 0.029x. This represents $y = \ln(m)$ as a function of $x = t$.

For each option, define a function for $m$ in terms of $x$ (time) and then graph its natural log:

• For A: type g(x) = ln(4.6 - 0.029x).
• For B: type h(x) = ln(e^(4.6 - 0.029x)).
• For C: type k(x) = ln(4.6*(0.971)^x).
• For D: type p(x) = ln(e^(-4.6 + 0.029x)).

Look at which of the new graphs of $\ln(m)$ lies exactly on top of the original line y = 4.6 - 0.029x for all $x$ where it’s defined. The corresponding $m$-formula is the correct answer.

The problem says the chemist graphs $\ln(m)$ versus $t$ and the points lie near the line

That means this equation is a model for the natural logarithm of the mass, not for the mass $m$ itself.

We want an equation with $m$ alone as a function of $t$. Right now we have $\ln(m)$ on the left. The natural log function $\ln(x)$ and the exponential function $e^x$ are inverses, so applying $e^{\,\cdot}$ to both sides will undo the $\ln$ on the left.

Start from

and apply the exponential with base $e$ to both sides of the equation:

Exponentiating both sides gives

Because $e^{\ln(m)} = m$, this simplifies to

This matches choice B, so the correct model is $m=e^{4.6-0.029t}$.