Problem 61 - College Algebra and Calculus: An Applied Approach

Question

In Exercises, solve for \(x\) or \(t\). $$ 300 e^{-0.2 t}=700 $$

Step-by-Step Solution

Answer

The solution to the equation \(300 e^{-0.2 t}=700\) for \(t\) is \[t= \frac{\ln\left(\frac{700}{300}\right)}{-0.2}\].

Step 1: Isolate exponential term

The first step involves isolating the exponential term, \(e^{-0.2 t}\). This can be achieved by dividing both sides of the equation by 300. So,\[ e^{-0.2 t}=\frac{700}{300} \]

Step 2: Applying the natural logarithm

A next step in solving such exponential equations is applying the natural logarithm to both sides of the equation from previous step. Using the property that \(\ln(e^x) =x\), the equation becomes: \[-0.2 t = \ln\left(\frac{700}{300}\right)\]

Step 3: Solving for \(t\)

Now, isolate \(t\) by dividing both side of the equation by \(-0.2\). Hence, the value of \(t\) is: \[t= \frac{\ln\left(\frac{700}{300}\right)}{-0.2}\]

Key Concepts

Natural Logarithm

In the world of mathematics, the natural logarithm is a fundamental concept used in various fields such as calculus and exponential equations. The natural logarithm, denoted as \( \ln \), uses the base \( e \), where \( e \) is approximately equal to 2.71828. This number is significant because it represents continuous growth. When solving exponential equations, the natural logarithm is extremely useful because it allows us to transform multiplicative relationships into additive ones, making the equations easier to solve.

• Any number raised to the power \( x \) can be expressed in a logarithmic form: \( x = \ln(b) \), where \( b = e^x \).
• Using the natural logarithm can simplify the process of solving equations involving the exponent \( e \).

In our example, \( -0.2t = \ln\left(\frac{700}{300}\right) \) was achieved by applying the natural logarithm to both sides of the equation. This was possible because of the property \( \ln(e^x) = x \), which "cancels out" the exponential expression, leaving us with a linear equation.

Isolating Variables

Isolating variables is a critical step in solving algebraic equations. This process involves arranging the equation so that the variable of interest is by itself on one side of the equation. This makes it easier to determine the value of the variable.

• Start with the original equation and identify the term containing the variable.
• Use algebraic operations such as addition, subtraction, multiplication, or division to "isolate" the variable.

In the given problem, the variable \( t \) must be isolated to find its value. Initially, the exponential term \( e^{-0.2t} \) was isolated by dividing both sides of the equation \( 300 e^{-0.2 t} = 700 \) by 300. After this step, further operations helped isolate \( t \) until we could solve for its numerical value. Isolating variables is a foundational technique that simplifies the solving process by gradually making the variable the main focus of the equation.

Solving Algebraic Equations

Algebraic equations often require a methodical approach to reach a solution. Solving involves finding the value of unknown variables that make the equation true. Here's how to effectively solve such equations:

• Start by simplifying the equation, if possible.
• Combine like terms and ensure all similar terms are on the same side.
• Use properties of equality to perform operations equally on both sides of the equation.

In the example at hand, each step focused on systematically reducing the complexity of the equation. We started by isolating the exponential expression, then applied the natural logarithm to transition to a linear equation, and finally isolated the variable \( t \) to compute its value. By the time we reached \( t = \frac{\ln\left(\frac{700}{300}\right)}{-0.2} \), each step had brought us closer to the solution in a clear, logical manner. Such a systematic approach is key to solving even the most complex algebraic equations efficiently.