Problem 9
Question
Show that \(y=A B^{t}\) with \(B<1\) has a half-life of $$ T_{\text {Half }}=\frac{\log \frac{1}{2}}{\log B}=\frac{-\log 2}{\log B} $$
Step-by-Step Solution
Verified Answer
The half-life is \( T_{\text{Half}} = \frac{-\log 2}{\log B} \).
1Step 1: Understanding the Half-Life Concept
The half-life of a function describes the time it takes for a quantity to reduce to half its initial value. For the given function \( y = A B^t \), the goal is to find \( T_{\text{Half}} \) such that \( y(t=T_{\text{Half}}) = \frac{A}{2} \).
2Step 2: Substitute Half-Life Condition
We know that at \( t = T_{\text{Half}} \), \( y = \frac{A}{2} \). Substitute into the equation: \[ \frac{A}{2} = A B^{T_{\text{Half}}} \].
3Step 3: Simplify the Equation
Cancel \( A \) from both sides of the equation (assuming \( A eq 0 \)): \[ \frac{1}{2} = B^{T_{\text{Half}}} \].
4Step 4: Take the Logarithm
Take the logarithm on both sides to bring down the exponent: \[ \log \frac{1}{2} = T_{\text{Half}} \cdot \log B \].
5Step 5: Solve for \(T_{\text{Half}}\)
Rearrange the equation from the previous step to solve for \( T_{\text{Half}} \): \[ T_{\text{Half}} = \frac{\log \frac{1}{2}}{\log B} = \frac{-\log 2}{\log B} \].
6Step 6: Final Review
We have shown the expression for the half-life of \( y = A B^t \) with \( B < 1 \) as \( T_{\text{Half}} = \frac{-\log 2}{\log B} \). This matches the formula given in the problem statement.
Key Concepts
Half-LifeLogarithmsExponential Functions
Half-Life
The concept of half-life is quite fascinating and crucial in understanding how quantities decrease over time. Imagine you have something that decays or decreases systematically, like a radioactive substance or a medication in the bloodstream. The half-life is the time it takes for half of that substance to disappear or be used up. For this reason, it is widely used in physics, biology, and chemistry to describe various processes.
In the given exercise, the function is defined as \(y = A B^t\). The half-life \(T_{\text{Half}}\) is the time at which the quantity \(y\) becomes half of its initial value \(A\). That means, at \(t = T_{\text{Half}}\), \(y = \frac{A}{2}\). This is an important step because it directly sets the relationship for calculating half-life based on the decay factor \(B\).
Knowing that \(B < 1\), you're essentially looking at a decay process. Each time period, the quantity decreases by a constant proportion, showing the beauty and predictability of exponential decay.
In the given exercise, the function is defined as \(y = A B^t\). The half-life \(T_{\text{Half}}\) is the time at which the quantity \(y\) becomes half of its initial value \(A\). That means, at \(t = T_{\text{Half}}\), \(y = \frac{A}{2}\). This is an important step because it directly sets the relationship for calculating half-life based on the decay factor \(B\).
Knowing that \(B < 1\), you're essentially looking at a decay process. Each time period, the quantity decreases by a constant proportion, showing the beauty and predictability of exponential decay.
Logarithms
Logarithms are the magic key to solving problems involving exponential growth or decay. They help by converting multiplicative processes into additive ones, making it easier to solve equations where the unknown is an exponent.
When you have an equation like \(\frac{1}{2} = B^{T_{\text{Half}}}\), taking the logarithm allows you to bring the exponent, \(T_{\text{Half}}\), down in front of the logarithm, meaning \(\log \frac{1}{2} = T_{\text{Half}} \cdot \log B\).
When you have an equation like \(\frac{1}{2} = B^{T_{\text{Half}}}\), taking the logarithm allows you to bring the exponent, \(T_{\text{Half}}\), down in front of the logarithm, meaning \(\log \frac{1}{2} = T_{\text{Half}} \cdot \log B\).
- This step is crucial, as it takes a seemingly complex variable problem and turns it into a linear one.
- Logarithms with different bases can be converted using the change of base formula, allowing flexibility.
Exponential Functions
Exponential functions are at the heart of many natural processes and this exercise revolves around understanding their properties. Specifically, with \( y = A B^t \), where \(B < 1\), we're observing a decay over time, as opposed to growth (which you'd have if \(B > 1\)).
The form \(A B^t\) implies that each unit increase in \(t\) results in the multiplication of \(y\) by the constant \(B\). If \(B < 1\), this is known as exponential decay. This characteristic tells us that as time progresses, \(y\) will decrease, eventually reaching half its initial value by \(T_{\text{Half}}\).
The form \(A B^t\) implies that each unit increase in \(t\) results in the multiplication of \(y\) by the constant \(B\). If \(B < 1\), this is known as exponential decay. This characteristic tells us that as time progresses, \(y\) will decrease, eventually reaching half its initial value by \(T_{\text{Half}}\).
- Exponential functions showcase a constant percentage decay per time unit.
- They are used in real-world scenarios like population decline, radioactive decay, and cooling processes.
Other exercises in this chapter
Problem 7
Light intensities, \(I_{1}\) and \(I_{2}\), are measured at depths \(d\) in meters in two lakes on two different days and found to be approximately $$ I_{1}=22^
View solution Problem 8
a. The mass of a single \(V\). natriegens bacterial cell is approximately \(210^{-11}\) grams. If at time 0 there are \(10^{8} \mathrm{~V}\). natriegens cells i
View solution Problem 7
The nitrogen partial pressure in a muscle of a scuba diver is initially 0.8 atm. She descends to 30 meters and immediately the \(\mathrm{N}_{2}\) partial pressu
View solution