Iodine half life rewritten in hours

The model works in days: $h(d)=A_0(0.5{)}^{d/8}$. Time in hours is $t=24d$, so $d=\frac{t}{24}$. Substitute that for $d$: $$k(t)=A_0(0.5{)}^{\frac{t/24}{8}}=A_0(0.5{)}^{\frac{t}{192}}$$ because $24\cdot 8=192$. The half-life of 8 days is $192$ hours, and the exponent $\frac{t}{192}$ counts how many half-lives have passed.

None of the choices is written exactly like that, so rewrite with the power of a power rule, $(a^m{)}^n=a^{mn}$: $$(0.5{)}^{t/192}={\left({0.5}^{1/192}\right)}^t$$ That is choice D.

A check at one half-life: $t=192$ gives ${\left({0.5}^{1/192}\right)}^{192}=0.5$, so half the sample remains, as it should.

Each choice

• A. This uses 24 hours as the half-life. At $t=24$ it gives half the sample left after one day, but the half-life is 8 days. The 8 in the original exponent has been dropped.
• B. Simplified, this is ${0.5}^{8t/24}={0.5}^{t/3}$, a half-life of 3 hours. Here 8 multiplies $t$ where it should divide it: the exponent needs $\frac{t}{24\cdot 8}$, not $\frac{8t}{24}$.
• C. Simplified, this is ${0.5}^{24t/8}={0.5}^{3t}$, so a third of an hour would be a half-life. The conversion went the wrong way: hours were multiplied by 24 rather than divided by 24 to get days.
• D (correct). Replacing $d$ with $\frac{t}{24}$ gives ${0.5}^{t/192}$, and the power rule rewrites that as ${\left({0.5}^{1/192}\right)}^t$.