91Ó°ÊÓ

Natural logarithms are used to solve equations involving exponential terms where the base of the exponent is the natural number, known as Euler's number (\( e \approx 2.718 \)). In our problem, we deal with equations such as \( e^{-2t} \), where the exponent includes the variable we need to solve for. The natural logarithm, denoted as \( \ln \), is the inverse operation of the exponential function with base \( e \).
Here's how it works:

• To solve for a variable in an exponent, take the natural logarithm of both sides of the equation.
• Use the rule: \( \ln(e^x) = x \).
• This step simplifies the expression, allowing you to solve for the variable.

In the example solution, we took \( \ln \) of both sides in steps involving \( e^{-2t} = \frac{1}{3} \) and \( e^{-2t} = \frac{1}{9} \). This transformation turned the exponential equation into a linear one, solving for \( t \).
Utilizing natural logarithms makes dealing with exponential equations straightforward and manageable when solving for input variables within exponents.

When solving for the input variable, \( t \), in this type of equation, you'll often follow a series of clear steps. The equation in our example is \( s(t) = \frac{120}{1+3e^{-2t}} \), and we need to find \( t \) such that \( s(t) = 60 \) or \( s(t) = 90 \).
The strategy involves:

• Setting the equation: Start by replacing \( s(t) \) with the desired output value.
• Clearing fractions: Multiply both sides by terms that eliminate the denominator.
• Simplifying expressions: Distribute and combine like terms.
• Isolating the exponential term: Get \( e^{-2t} \) by itself on one side of the equation.
• Using logarithms: Apply natural logarithms to solve for \( t \).

Each step makes the original equation simpler, eventually revealing the expression for the variable within the exponent. Following these methods helps in systematically finding the solution for the input variable.

Fractional equations involve variables in the denominators that need to be carefully managed. Our initial function \( s(t) = \frac{120}{1+3e^{-2t}} \) is a classic fractional equation and needs a particular solving approach, especially when setting it equal to specific values like 60 and 90.
Here's the breakdown:

• To "clear" the fraction, multiply every term by the denominator \( 1 + 3e^{-2t} \).
• This action removes the denominator from the equation, simplifying it to a more manageable form.
• It allows the focus to shift to solving the polynomial equation left behind.

Clearing fractions is an essential initial step in many algebraic processes. It serves to streamline complex equations into simpler forms, making subsequent calculations easier and more organized.