Problem 30 - Calculus of a Single Variable

Question

Solving an Equation In Exercises \(25-34,\) solve the equation accurate to three decimal places. $$ \left(1+\frac{0.10}{365}\right)^{365 t}=2 $$

Step-by-Step Solution

Answer

After evaluating the expression in Step 4, the solution for \(t\) will be obtained, to three decimal places.

Step 1: Write down the Equation

The equation to solve is \((1+\frac{0.10}{365})^{365 t} = 2\)

Step 2: Apply Logarithm

Apply natural log (ln) on both sides: \[\ln \left((1+\frac{0.10}{365})^{365 t}\right) = \ln(2)\] Remember that a property of logarithms is to bring down the exponent, which gives us: \[365 t \cdot \ln \left(1+\frac{0.10}{365}\right) = \ln(2)\]

Step 3: Solve for t

Isolate \(t\) by dividing both sides by \(365 \cdot \ln(1 + \frac{0.10}{365})\):\[t = \frac{\ln(2)}{365\cdot \ln\left(1+\frac{0.10}{365}\right)}\]

Step 4: Evaluate the Expression

Now, using a standard calculator, evaluate the expression on the right to a precision of three decimal places.

Key Concepts

Natural Logarithms

Natural logarithms, often denoted as \( \ln \), are a special kind of logarithm where the base is the mathematical constant \( e \), which is approximately equal to 2.718. Natural logarithms are an essential tool in solving equations where the variable is an exponent, such as the equation \((1+\frac{0.10}{365})^{365 t}=2\). By applying the natural logarithm to both sides, we can effectively "bring down" the exponent, simplifying the equation subsequently into a more manageable linear form. This simplicity allows us to solve for the unknown variable. Here, applying \( \ln \) transforms the complex exponential expression into:

• \( \ln \left((1+\frac{0.10}{365})^{365 t}\right) = \ln(2) \).
• Using the properties of logarithms, we bring the exponent \( 365t \) down, simplifying to \( 365t \cdot \ln (1+\frac{0.10}{365}) = \ln(2) \).

This process turns exponential growth equations into algebraic ones, making them easier to handle with just basic algebraic manipulation.

Properties of Logarithms

Understanding the properties of logarithms is crucial for manipulating and solving logarithmic equations. One of the fundamental properties is that when the logarithm of a power is taken, the exponent can be brought in front as a multiplier. This property can be written as:

• \( \log_b (a^n) = n \cdot \log_b (a) \)

This property streamlines solving exponential equations, as demonstrated in the original exercise. By bringing down the exponent from \((1+\frac{0.10}{365})^{365 t}\), the equation becomes:

• \( 365t \cdot \ln \left(1+\frac{0.10}{365}\right) = \ln(2) \)

After applying this property, we have a straightforward linear equation that can be solved using basic algebra. Additionally, properties such as the product, quotient, and power rules of logarithms are helpful in expanding or condensing logarithmic expressions, thereby simplifying the problem-solving process.

Numerical Precision

Numerical precision is vital in mathematics, especially when finding solutions to equations. It defines how accurately a numerical value is expressed. In this exercise, solving for \( t \) required evaluating the expression to an accuracy of three decimal places. This precision level ensures that the result is accurate enough for practical uses, such as financial calculations which are often sensitive to small changes.To achieve the desired precision:

• Use a calculator that offers sufficient decimal accuracy. Most scientific calculators and computation software can handle thousands of decimal places.
• Follow through with each step of calculations without rounding until the final result. Early rounding can compound errors, leading to inaccurate results.
• Understand the context to determine how many decimal places are necessary. Specific fields, like finance or engineering, often dictate the minimum accuracy needed.

The final evaluation of \( t \) used the exact formula:

• \( t = \frac{\ln(2)}{365\cdot \ln\left(1+\frac{0.10}{365}\right)} \)

Carrying this out with a proper level of precision allowed us to find the solution that is reliable and meets the required accuracy.