Recipe

Find when a decaying drug concentration reaches a given level

Use when: you know $C_0$ , the elimination constant $k_e$ , and a target concentration — including the special target $C_0/2$ , the half-life

Goal: the exact time $t$ at which $C(t)$ first reaches that target level

Taught in Pharmacokinetics: a drug-elimination model

Quick recipe

Write down the exponential decay model with the known constants
Set $C(t)$ equal to the target level and isolate the exponential
Take $\ln$ of both sides and solve for $t$
Sanity-check the answer's size against the half-life

1
Start from $C(t)=C_0e^{-k_et}$ , substituting the known $C_0$ and $k_e$ .

Example

$C_0=200$ mg/L, $k_e=0.2$ /hour: $C(t)=200e^{-0.2t}$ .

Why: every question in this section reduces to solving this one equation for whichever quantity is unknown.
2
Divide both sides by $C_0$ so the exponential stands alone.

Example

Target $=25$ mg/L: $200e^{-0.2t}=25 \Longrightarrow e^{-0.2t}=\dfrac{25}{200}=0.125$ .
- the target is exactly $C_0/2$ — this is the half-life question; Step 3's shortcut applies directly.
- the target is any other level — continue to Step 3 the general way.
Why: $\ln$ can only be applied cleanly once the exponential is isolated on one side.
3
$\ln(e^{-k_et})=-k_et$ , so $t=\dfrac{-\ln(\text{ratio})}{k_e}=\dfrac{\ln(C_0/\text{target})}{k_e}$ .

Example

$t=\dfrac{-\ln(0.125)}{0.2}=\dfrac{\ln 8}{0.2}=\dfrac{2.079}{0.2}\approx10.40$ hours.

⚡ Shortcut

When the target is exactly $C_0/2$ (the half-life question), $\ln(C_0/\text{target})=\ln2$ every time, regardless of $C_0$ — so $t_{1/2}=\dfrac{\ln2}{k_e}$ can be read off directly, with no division by $C_0$ needed at all. For $k_e=0.2$ : $t_{1/2}=\dfrac{0.6931}{0.2}\approx3.47$ hours.

Why: this is ordinary log-solving, but the sign is easy to lose — $-k_et$ is negative and $\ln(\text{ratio})$ is negative too (the ratio is under $1$ ), and the two negatives cancel to give a positive $t$ .
4
Compare the computed $t$ to $t_{1/2}=\ln2/k_e$ : a target above $C_0/2$ should give $t<t_{1/2}$ , and a target below $C_0/2$ should give $t>t_{1/2}$ .

Example

$t_{1/2}\approx3.47$ hours; the target $25$ is below $C_0/2=100$ , so $t$ should exceed $3.47$ — and indeed $10.40>3.47$ , confirming the answer is at least plausible.

Why: this catches a sign error or a misplaced ratio before it's reported as the final answer.

Common mistakes

Taking $\ln$ before dividing by $C_0$ , mishandling $\ln(C_0)-k_et=\ln(\text{target})$ instead of isolating the exponential first.
Forgetting the negative sign on $k_et$ , or flipping the final division, and reporting a negative time.
Using the half-life shortcut for a target that isn't $C_0/2$ — the shortcut only works because $\ln(C_0/(C_0/2))=\ln2$ is a constant; any other ratio needs the general Step 3.
Stopping at the ratio $C(t)/C_0$ instead of continuing on to solve for $t$ itself.