Question

Geometric means A quantity grows exponentially according to \(y(t)=y_{0} e^{k t} .\) What is the relationship among \(m, n,\) and \(p\) such that \(y(p)=\sqrt{y(m) y(n)} ?\)

Short answer

Answer: The relationship between the time points \(m\), \(n\), and \(p\) is given by the equation \(2kp= km+kn\).

Step-by-step solution

Write down the given quantities and equation

We are given the exponential growth function \(y(t)=y_0e^{kt}\), where \(y_0\) is the initial quantity, \(k\) is the growth rate, and \(t\) is time. We are asked to find the relationship between the time points \(m\), \(n\), and \(p\) such that $$y(p)=\sqrt{y(m)y(n)}.$$

Substitute the exponential growth function in the equation

We can substitute the given exponential growth function into the equation to obtain the following: $$y_0e^{kp}=\sqrt{y_0e^{km}y_0e^{kn}}.$$

Simplify the equation

Since \(y_0 > 0\), we can safely divide both sides of the equation by \(y_0\) without changing the equality. Let's do this and then simplify further: $$e^{kp}=\sqrt{e^{km}e^{kn}}.$$Using the rule for combining exponents, we get:$$e^{kp}=\sqrt{e^{(km+kn)}}.$$

Take the natural logarithm of both sides

To find the relationship between the exponents \(km\), \(kn\), and \(kp\), let's take the natural logarithm of both sides of the equation:$$\ln{e^{kp}}=\ln{\sqrt{e^{(km+kn)}}}.$$Recall that \(\ln{e^x}=x\). So, we have:$$kp=\frac{1}{2}(km+kn).$$

Write the final relationship

Rearranging the equation, we get the final relationship among the time points \(m\), \(n\), and \(p\) such that the given property holds:$$2kp= km+kn.$$This is the relationship among \(m\), \(n\), and \(p\) as described in the exercise.

Key concepts

Geometric Mean

The geometric mean is a useful way to determine the average growth rate of quantities. Unlike the arithmetic mean, which adds numbers together, the geometric mean multiplies them and then takes the root of the result. Specifically, if you have two numbers, the geometric mean is the square root of their product. It's calculated as \(\sqrt{ab}\) for two quantities \(a\) and \(b\). - The geometric mean is particularly useful in the context of exponential growth or decay because its application aligns with the multiplicative nature of these processes. - In our exercise, the task was to relate the exponential function values at different times \(m\), \(n\), and \(p\) using geometric mean.- By setting an exponential growth equation's value at time \(p\) equal to the geometric mean of its values at \(m\) and \(n\), hybrid arithmetic was used to find a meaningful relationship between \(m\), \(n\), and \(p\). This concept helps simplify and manage exponential growth modeling.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They commonly appear as \(y(t) = y_0 \cdot e^{kt}\), where:- \(y_0\) represents the initial amount or starting value.- \(e\) is Euler's number, an irrational constant approximately equal to 2.71828.- \(k\) denotes the growth or decay rate.- \(t\) is time, often treated as a variable.Exponential functions are used in many fields like population growth, radioactive decay, and finance. What makes these functions powerful is their ability to model rapid changes, both growth, and decay, over time. - As the variable \(t\) increases, the value of \(y(t)\) grows or shrinks exponentially, meaning at rapidly increasing or decreasing rates.- In our exercise, substituting the exponential function revealed the parts of the expression controlled by time points \(m\), \(n\), and \(p\). Understanding these functions allows us to dissect the problem and find the relationships between different parameters involved in exponential changes.

Natural Logarithms

Natural logarithms, often symbolized as \(\ln\), provide insight into exponential equations by effectively "undoing" them. The natural logarithm of a number is the power to which \(e\) must be raised to equal that number. Essentially, when you see \(\ln(x)\), you're asking: "What power of \(e\) equals \(x\)?"- The natural logarithm is uniquely suited to working with exponential functions, allowing us to linearize problems involving exponentials.- For instance, if you have an equation like \(e^x = y\), the natural logarithm lets you rewrite it as \(x = \ln(y)\). This transformation is crucial for solving equations where the variable is an exponent.In the step-by-step solution, natural logarithms transformed the problem into a format where the relationship between \(kp\), \(km\), and \(kn\) could be more clearly understood. By taking the natural log of both sides, we transformed a multiplicative relationship into an additive one, thus simplifying our understanding and solution of the equation at hand. Remember, using natural logarithms is a common practice when dealing with exponential functions, especially in solving for unknown exponents.