Problem 43 - Precalculus with Limits

Question

Solve the exponential equation algebraically. Approximate the result to three decimal places. \(\left(1+\frac{0.065}{365}\right)^{365 t}=4\)

Step-by-Step Solution

Answer

The solution for \(t\) is approximately 10.646 when rounded to three decimal places.

Step 1: Transform Equation

Transform the equation \(\left(1+\frac{0.065}{365}\right)^{365t}=4\) using basic logarithm rules. The first step is to log both sides of the equation using the natural logarithm. We will use the identity \(\ln(a^b) = b \ln(a)\): \(\ln((1+0.065/365)^{365t})=\ln(4)\)

Step 2: Simplify Equation

Applying the logarithm rule we get: \(365t \cdot \ln\left(1+\frac{0.065}{365}\right)=\ln(4)\)

Step 3: Solve for \(t\)

Isolate \(t\) by first dividing both sides of the equation by \(365\):\(t \cdot \ln\left(1+\frac{0.065}{365}\right) = \frac{\ln(4)}{365}\), and then dividing both sides by \(\ln\left(1+\frac{0.065}{365}\right)\) to solve for \(t\): \(t = \frac{\ln(4)}{365 \cdot \ln\left(1+\frac{0.065}{365}\right)}\)

Step 4: Compute Numerical Value

Compute numerical value from given equation and round to three decimal places: \(t \approx 10.646\)

Key Concepts

Logarithmic Function

Logarithmic functions are the inverses of exponential functions. They help us solve equations where the unknown is an exponent, as in our given exercise. The basic idea is to "undo" an exponent by applying a logarithm, which is how we simplify the problem to find the value of the unknown.

• Using logarithms, an equation like \(a^b = c\) can be rewritten using the logarithmic function as \(b = \log_a(c)\).
• This process makes it easier to isolate the variable, especially when dealing with complex or large numbers.
• In mathematical terms, when we apply the natural logarithm (ln) to both sides of a given equation, we utilize the identity \(\ln(a^b) = b \ln(a)\). This rule simplifies the exponent into a multiplication term, which is easier to work with algebraically.

Understanding and applying this technique are crucial for solving exponential equations like the one in our exercise. It's a tool that converts an exponential relationship into something more linear and manageable for solving.

Natural Logarithm

The natural logarithm is a specific type of logarithm that uses the mathematical constant \(e\) (approximately 2.718) as its base. It is denoted by \(\ln(x)\). Natural logarithms are widely used in mathematics and many scientific applications due to their nice analytical properties.

• The natural logarithm allows us to transform exponential growth and decay problems into simple linear scenarios by taking advantage of its convenient properties.
• In our exercise, we used \(\ln(x)\) because it makes differentiation and integration straightforward due to its boundary property \(\lim_{x \to \infty} \frac{\ln(x)}{x} = 0\).
• The exponential and logarithmic functions are inverses of each other, meaning \(\ln(e^x) = x\) and \(e^{\ln(x)} = x\). This inverse relationship is fundamental when manipulating equations like the one we are solving.

Thus, using the natural logarithm helped us handle the original exponential expression efficiently and solve for \(t\) step by step.

Numerical Approximation

Numerical approximation is a technique used to find approximate solutions to mathematical problems that may not have exact answers or where exact answers are inconvenient. It is especially useful when a precise numerical value is necessary.

• In the context of exponential equations, after transforming and simplifying the problem using logarithms, the next step is to compute a numerical value.
• Since exact solutions can be challenging or impossible to achieve due to irrational numbers, approximations give us the next best thing. The key is to ensure the approximation is accurate enough for practical purposes.
• For example, the solution \(t \approx 10.646\) involves calculating \(t\) using a calculator, rounding the decimal to three places for clarity and applicability, ensuring any real-world computations or applications of \(t\) remain reliable.

Numerical approximation adds value in scenarios where exact arithmetic solutions aren't feasible, thus serving a crucial role in both theoretical and applied mathematics.