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)^{(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) \\ (P(640))/(P_0)=0.47 \end{gathered}`

Now simply substitute this ratio into the above equation

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

Take log on both sides

`\begin{gathered} \ln (0.47)=\ln ((0.5)^{(640)/(H)}) \\ \ln (0.47)=(640)/(H)\ln (0.5) \\ (\ln (0.47))/(\ln (0.5))=(640)/(H) \\ 1.0893=(640)/(H) \\ H=(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)^{(t)/(H)} \\ 55=100\cdot(0.5)^{(t)/(588)} \\ (55)/(100)=0.5^{(t)/(588)} \\ 0.55=0.5^{(t)/(588)} \end{gathered}`

Take log on both sides

`\begin{gathered} \ln (0.55)=\ln (0.5^{(t)/(588)}) \\ \ln (0.55)=(t)/(588)\ln (0.5) \\ (\ln (0.55))/(\ln (0.5))=(t)/(588) \\ 0.8625=(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.