Question

I'm having a problem with these exponential and logarithmic functions I will include picture

Answer 1

The half-life of a radioactive element is given by

$P(t)=P_0\cdot(0.5)^{\frac{t}{H}}$

Where P(t) is the amount of element at time t, P₀ is the amount at time zero.

H is the half-life and t is the time in days.

For the given case,

t = 640 days

$\begin{gathered} P\mleft(640\mright)=P_0\cdot(1-0.53) \\ P(640)=P_0\cdot(0.47) \\ \frac{P(640)}{P_0}=0.47 \end{gathered}$

Now simply substitute this ratio into the above equation

$\begin{gathered} \frac{P(t)}{P_0}=(0.5)^{\frac{t}{H}} \\ 0.47=(0.5)^{\frac{640}{H}} \end{gathered}$

Take log on both sides

$\begin{gathered} \ln (0.47)=\ln ((0.5)^{\frac{640}{H}}) \\ \ln (0.47)=\frac{640}{H}\ln (0.5) \\ \frac{\ln (0.47)}{\ln (0.5)}=\frac{640}{H} \\ 1.0893=\frac{640}{H} \\ H=\frac{640}{1.0893} \\ H=587.5 \\ H=588\; \text{days} \end{gathered}$

Therefore, the half-life of the element is 588 days

Part (b)

How long will it take for a sample of 100 mg to decay to 55 mg

So, we need to find time (t)

For the given case, we have

P(t) = 55

P₀ = 100

H = 588

$\begin{gathered} P(t)=P_0\cdot(0.5)^{\frac{t}{H}} \\ 55=100\cdot(0.5)^{\frac{t}{588}} \\ \frac{55}{100}=0.5^{\frac{t}{588}} \\ 0.55=0.5^{\frac{t}{588}} \end{gathered}$

Take log on both sides

$\begin{gathered} \ln (0.55)=\ln (0.5^{\frac{t}{588}}) \\ \ln (0.55)=\frac{t}{588}\ln (0.5) \\ \frac{\ln (0.55)}{\ln (0.5)}=\frac{t}{588} \\ 0.8625=\frac{t}{588} \\ t=588\cdot0.8625 \\ t=507\; \text{days} \end{gathered}$

Therefore, it would take 507 days for a sample of 100 mg to decay to 55 mg.