Question

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

Short answer

The solution is \(t\approx10.539\)

Step-by-step solution

Write the equation

The equation which needs to be solved is \(\left(1+\frac{0.065}{365}\right)^{365 t}=4.\)

Applying the natural logarithm

First, apply natural logarithm to both sides of the equation. This helps to simplify the term in the exponent. We have \(\ln\left(\left(1+\frac{0.065}{365}\right)^{365 t}\right)=\ln(4)\). Since \(\ln(a^b)=b\ln(a)\), the equation simplifies to \(365t\ln\left(1+\frac{0.065}{365}\right)=\ln(4)\)

Solve for the variable

Rearrange the equation for \(t\). Therefore, \(t=\frac{\ln(4)}{365\ln\left(1+\frac{0.065}{365}\right)}\)

Numerical approximation

Using a calculator or similar tools, approximate the value of \(t\) up to three decimal places: \(t\approx 10.539 \)

Key concepts

Natural Logarithm

The natural logarithm is a fundamental concept used to solve exponential equations. It involves representing a number as the power to which the base of the natural log, which is Euler's number \( e \), must be raised to equal that number. When solving exponential equations, applying the natural logarithm can greatly simplify the process because of its unique properties.

• Simplification: An exponent in an equation becomes a coefficient when a logarithm is applied, turning a complex exponential equation into a linear one.
• Logarithm Rule: The logarithm of a power \(\ln(a^b) = b\ln(a)\) allows you to move the exponent to the front, making it easier to solve for the variable.
• Application: In this equation, applying \(\ln\) to both sides allowed us to bring \(365t\) down from the exponent, facilitating further algebraic manipulation.

Understanding the natural logarithm is crucial, as it is a powerful tool in calculus and higher-level mathematics, particularly in solving exponential and logarithmic equations.

Algebraic Solutions

Solving exponential equations using algebraic methods involves manipulating equations to isolate the variable. The goal is usually to simplify the equation such that the unknown variable is easy to extract. In this scenario:

• Equation Setup: Start by identifying the form of the exponential equation, as given by \(\left(1+\frac{0.065}{365}\right)^{365t}=4\).
• Use of Logarithms: We applied the natural logarithm to both sides to help remove the exponent, rearranging the terms algebraically.
• Isolating the Variable: The goal was to isolate \( t \) by dividing both sides by the constants and expressions not involving \( t \).
• Result: By following these steps, we obtain a clear expression for \( t \), \( t = \frac{\ln(4)}{365\ln\left(1+\frac{0.065}{365}\right)} \), showing the power of algebraic manipulation in finding solutions.

Algebraic solutions for exponential equations often require a good grasp of properties of exponents and logarithms, as these are frequently used steps in simplification.

Numerical Approximation

Once you have derived an expression for the variable, numerical approximation helps determine an actual numerical value. Numerical approximation is essential when an equation cannot be solved exactly or to make results more interpretable in real-world contexts. Here's how we tackled this:

• Calculator Use: Often, calculators or computational tools are used to obtain precise decimal approximations, as seen with \( t \approx 10.539 \).
• Why Approximation: Many solutions involve irrational numbers that can't be expressed exactly in decimal form, hence the need to approximate to a set number of decimal places.
• Precision: Approximating to three decimal places is common in applied mathematics as it provides a balance between precision and simplicity, especially in scientific measurements.
• Application: In cases involving exponential growth or decay, such as interest calculations as reflected in compounded interest problems, precise numerical approximations can significantly impact the final result.

Grasping numerical approximation is crucial for solving real-world problems where exact numbers might not be possible or practical, making it a key skill in fields ranging from finance to engineering.