Exponential Decay: Finding a Time

Exponential decay models work like growth, with a negative exponent. Isolate the exponential, take $\ln$, then divide, the signs work out to a positive time because the ratio is less than $1$.

The mass $N$ (in grams) of a decaying substance after $t$ years is $N = 100e^{-0.05t}$. Find, to the nearest year, the time for the mass to halve to $50$ g. [4]

The working

Step 1, set $N = 50$ and isolate the exponential: $50 = 100e^{-0.05t} \;\Rightarrow\; e^{-0.05t} = 0.5 \quad \text{(M1)}$

Step 2, take natural logs: $-0.05t = \ln 0.5 = -0.6931\ldots \quad \text{(M1)}$

Step 3, solve for $t$ (negative over negative gives positive): $t = \frac{-0.6931}{-0.05} = 13.86\ldots \quad \text{(A1)}$

To the nearest year, the mass halves after $14$ years. (A1)

Where the marks are won and lost

• Divide before taking logs: isolate $e^{-0.05t}$ first ($= 0.5$), then $\ln$.
• $\ln 0.5$ is negative ($\approx -0.693$), and dividing by $-0.05$ gives a positive $t$. Mishandling the double negative is the usual slip, the time must be positive.
• Use $\ln$ (base $e$) because the base is $e$.

Common mistakes

• Losing a negative sign and getting a negative (impossible) time.
• Using $\lg$ instead of $\ln$.
• Rounding $\ln 0.5$ too early.

Full method: e^x and ln x notes. Topic home: Logs & Exponentials pillar.