Problem 13 - Thomas Calculus

Question

In Exercises \(13-16,\) solve for \(t\). a. \(e^{-0.3 t}=27 \quad\) b. \(e^{k t}=\frac{1}{2} \quad\) c. \(e^{(\ln 0.2) t}=0.4\)

Step-by-Step Solution

Answer

a. \(t = \frac{\ln(27)}{-0.3}\), b. \(t = \frac{\ln\left(\frac{1}{2}\right)}{k}\), c. \(t = \frac{\ln(0.4)}{\ln(0.2)}\)."

Step 1: Solve for 't' in Equation A

The equation given is \(e^{-0.3t} = 27\). To solve for \(t\), we need to take the natural logarithm of both sides. This gives us: \(-0.3t = \ln(27)\). Now, solve for \(t\) by dividing both sides by \(-0.3\): \(t = \frac{\ln(27)}{-0.3}\).

Step 2: Solve for 't' in Equation B

The equation given is \(e^{kt} = \frac{1}{2}\). First, apply the natural logarithm to both sides to get \(kt = \ln\left(\frac{1}{2}\right)\). Then, solve for \(t\) by dividing both sides by \(k\): \(t = \frac{\ln\left(\frac{1}{2}\right)}{k}\).

Step 3: Solve for 't' in Equation C

The equation given is \(e^{(\ln 0.2) t} = 0.4\). Start by taking the natural logarithm of both sides: \((\ln 0.2) t = \ln(0.4)\). Now, solve for \(t\) by dividing both sides by \(\ln 0.2\): \(t = \frac{\ln(0.4)}{\ln(0.2)}\).

Key Concepts

Natural Logarithms

Natural logarithms are an essential mathematical tool for solving equations, especially those involving exponential functions. The natural logarithm, denoted by \( \ln \), is the logarithm to the base \( e \), where \( e \approx 2.71828 \). This unique constant, known as Euler's number, emerges naturally in various mathematical contexts, particularly those involving growth and decay processes.

When dealing with equations where the unknown variable is in the exponent, natural logarithms enable us to bring that variable down to a satisfactory position for easy solving. By applying the natural logarithm to both sides of an equation, we can dismantle the exponentiation, making it possible to isolate the variable. This property is especially useful for exponential equations like the ones in our exercise.

Remember, applying \( \ln \) to both sides of the equation is valid only if the original numbers are positive. This is because the domain of natural logarithmic functions is restricted to positive real numbers. Always check your values before executing this step!

Solving Equations

Solving equations can sometimes seem like a puzzle, but breaking it into straightforward strategies can make the task much easier. When an equation includes exponentials, natural logarithms can be used to effectively untangle and solve these equations.

Consider the steps outlined in the exercise:

• In Equation A, \( e^{-0.3t} = 27 \), the aim is to isolate \( t \). First, take the natural logarithm of both sides to get \(-0.3t = \ln(27)\), then solve for \( t \) by dividing by \(-0.3\).
• For Equation B, \( e^{kt} = \frac{1}{2} \), the process is similar. Take \( \ln \) of both sides to achieve \( kt = \ln\left(\frac{1}{2}\right)\), and isolate \( t \) by dividing by \( k \).
• Finally, Equation C, \( e^{(\ln 0.2) t} = 0.4 \), follows suit. Take natural logarithms to end up with \( (\ln 0.2) t = \ln(0.4) \), and solve for \( t \) with division.

As seen, the method relies heavily on the properties of logarithms to simplify exponential equations. Step-by-step, it allows clear isolation and solving of the variable. This logical approach keeps you organized and helps avoid mistakes.

Logarithmic Functions

Logarithmic functions, including natural logarithms, are powerful tools in mathematics. They have the inverse relationship with exponential functions, meaning they can "undo" what exponentiation does. This feature makes them indispensable when dealing with equations where the unknown is in an exponent, such as those seen in our problem sets.

By applying a logarithmic function, specifically the natural logarithm \( \ln \), we can reorganize exponential equations, turning a potentially complex scenario into a manageable one. In this process:

• The property \( \ln(e^{x}) = x \) allows any exponent containing \( e \) to be broken down to solve for \( x \).
• It is crucial in algebra and calculus, ensuring that exponential growth, decay, and other dynamic equations can be effectively modeled and solved.
• The logarithmic approach provides insight into the rate of change and the nature of continuous growth.

Hence, understanding how logarithmic functions operate not only helps in solving equations but also in comprehending a variety of scenarios where growth or decay processes are involved. These functions become essential in fields as diverse as biology, economics, and physics, making them a pivotal concept in advanced studies.