Problem 56 - Algebra and Trigonometry with Analytic Geometry

Question

The radioactive bismuth isotope \({ }^{210} \mathrm{Bi}\) disintegrates according to \(Q=k(2)^{-t / 5}\), where \(k\) is a constant and \(t\) is the time in days. Express \(t\) in terms of \(Q\) and \(k\).

Step-by-Step Solution

Answer

\( t = -5 \cdot \frac{\log\left(\frac{Q}{k}\right)}{\log(2)} \)."

Step 1: Identify the Given Formula

The given formula for disintegration is \( Q = k(2)^{-t/5} \). Our goal is to express \( t \) in terms of \( Q \) and \( k \).

Step 2: Isolate the Exponential Term

To isolate \((2)^{-t/5}\), first divide both sides of the equation by \( k \): \( \frac{Q}{k} = (2)^{-t/5} \).

Step 3: Apply Logarithm

Take the logarithm of both sides to solve for \(-t/5\): \( \log\left(\frac{Q}{k}\right) = \log\left((2)^{-t/5}\right) \).Using the property \( \log(a^b) = b \cdot \log(a) \), this becomes: \( \log\left(\frac{Q}{k}\right) = -\frac{t}{5} \cdot \log(2) \).

Step 4: Solve for \( t \)

Multiply both sides by \(-5 / \log(2)\) to solve for \( t \): \( t = -5 \cdot \frac{\log\left(\frac{Q}{k}\right)}{\log(2)} \).

Key Concepts

Radioactive Isotopes

Radioactive isotopes are fascinating substances found both in nature and in laboratories. These isotopes, like the bismuth isotope \( ^{210}\mathrm{Bi} \), possess an unstable nucleus. This instability leads to the emission of energy in the form of radiation over time.

• Each radioactive isotope has a unique half-life, which is the time it takes for half of the isotope to decay.
• The half-life is a crucial concept that helps scientists understand how quickly a substance will reduce in quantity.
• In equations, the decay is often represented with exponential functions, reflecting the ever-decreasing nature of the amount present over time.

Understanding these basics helps predict how long a radioactive isotope will effectively remain and be detectable.

Logarithms

Logarithms are essential in solving exponential equations, especially ones involved in radioactive decay. Essentially, a logarithm tells us the power to which a number— the base—must be raised to achieve another number. In simpler terms:

• If \( a^b = c \), then \( \log_a(c) = b \).
• This function is pivotal when trying to isolate a variable found in an exponent.
• With our exercise example, logarithms let us convert \( (2)^{-t/5} \) into a linear relationship with \( \log(2) \) acting as a constant scaler, making it much more straightforward to solve for \( t \).

Thus, using logarithms seamlessly cuts through the complexity of exponential relationships.

Algebraic Manipulation

Algebraic manipulation is at the heart of solving equations. It involves rearranging and simplifying equations to free a particular variable. In the context of radioactive decay:

• Start by rearranging the original equation to isolate the exponential term: \( \frac{Q}{k} = (2)^{-t/5} \).
• Next, apply logarithmic rules to both sides to simplify further: \( \log\left( \frac{Q}{k} \right) = -\frac{t}{5} \cdot \log(2) \).
• Ultimately, solve for the desired variable (\( t \)) by applying algebraic skills to multiply out constants, yielding the final result: \( t = -5 \cdot \frac{\log \left( \frac{Q}{k} \right)}{\log(2)} \).

These steps highlight how a systematic approach can transform complex expressions into solvable equations.