Question

Show mathematically that $2^{-t / T_{12}}=\left(\frac{1}{2}\right)^{t / T_{12}}=e^{-t / \tau}\( if and only if \)T_{1 / 2}=\tau \ln 2 .$ [Hint: Take the natural logarithm of each side. \(]\)

Short answer

Question: Show that \(2^{-t / T_{12}}=\left(\frac{1}{2}\right)^{t / T_{12}}=e^{-t / \tau}\) if and only if \(T_{1 / 2}=\tau \ln 2\). Answer: We showed that the given equation holds only if \(T_{12} = \tau \ln 2\) using logarithm properties and algebraic manipulations. Conversely, we proved that assuming \(T_{12} = \tau \ln 2\) implies the given equation. Hence, the given equation is true if and only if \(T_{1 / 2} = \tau \ln 2\).

Step-by-step solution

Proving that \(2^{-t / T_{12}}=\left(\frac{1}{2}\right)^{t / T_{12}}=e^{-t / \tau}\) implies \(T_{1 / 2}=\tau \ln 2\)

First, we'll prove the implication from the given equation to the desired result. Start with the given equation: \(2^{-t / T_{12}}=e^{-t / \tau}\) Now, take the natural logarithm of both sides: \(\ln{(2^{-t / T_{12}})} = \ln{e^{-t / \tau}}\) Using logarithm properties, we can simplify this to: \(-\frac{t}{T_{12}}\ln{2} = -\frac{t}{\tau}\) Now, cancel the \(-t\) term from both sides: \(\frac{\ln{2}}{T_{12}} = \frac{1}{\tau}\) Rearrange to solve for \(T_{12}\): \(T_{12} = \frac{\ln{2}}{1/ \tau} = \tau \ln{2}\) So, we have shown that the given equation implies \(T_{1 / 2}=\tau \ln 2\).

Proving that \(T_{1 / 2}=\tau \ln 2\) implies \(2^{-t / T_{12}}=\left(\frac{1}{2}\right)^{t / T_{12}}=e^{-t / \tau}\)

Now, we'll prove the implication from the desired result back to the given equation. Start with the given condition: \(T_{12} = \tau \ln 2\) Rewrite this as: \(\frac{1}{T_{12}} = \frac{1}{\tau \ln 2}\) Now, substitute this into the exponent of the equation we are trying to reach: \(2^{-t / T_{12}}=2^{-t / (\tau \ln 2)}\) Since \(\ln{(2^x)} = x \ln{2}\), we can rewrite this as: \(2^{-t / T_{12}}=e^{-t \frac{\ln{(2)}}{\tau \ln 2}}\) Now, cancel out the \(\ln 2\) in the exponent: \(2^{-t / T_{12}}=e^{-t / \tau}\) So, we have shown that \(T_{1 / 2}=\tau \ln 2\) implies \(2^{-t / T_{12}}=\left(\frac{1}{2}\right)^{t / T_{12}}=e^{-t / \tau}\). Thus, we have proved that \(2^{-t / T_{12}}=\left(\frac{1}{2}\right)^{t / T_{12}}=e^{-t / \tau}\) if and only if \(T_{1 / 2}=\tau \ln 2\).