Problem 302 - Precalculus

Question

Use logarithms to solve. \(2 e^{6 x}=13\)

Step-by-Step Solution

Answer

\(x \approx 0.227\)

Step 1: Isolate the Exponential Expression

Start by isolating the exponential term on one side of the equation. Divide both sides by 2 to get:\[e^{6x} = \frac{13}{2}\]

Step 2: Apply the Natural Logarithm

To solve for \(x\), take the natural logarithm (\(\ln\)) of both sides of the equation. Remember that the natural logarithm and the exponential function are inverses:\[\ln(e^{6x}) = \ln\left(\frac{13}{2}\right)\]

Step 3: Use the Logarithmic Identity

Utilize the logarithmic identity \(\ln(e^a) = a\) to simplify the left side of the equation:\[6x = \ln\left(\frac{13}{2}\right)\]

Step 4: Solve for x

To find \(x\), divide both sides of the equation by 6:\[x = \frac{\ln\left(\frac{13}{2}\right)}{6}\]Using a calculator, approximate this value if needed.

Key Concepts

Natural Logarithm

The natural logarithm, often expressed with the notation \( \ln \), is a fundamental concept when dealing with exponential equations. It is essentially the inverse operation to taking the exponential of a number, especially when dealing with base \( e \), which is approximately 2.718. This makes it particularly useful when solving equations where the variable is in an exponent.A key property of the natural logarithm is that it can transform exponential expressions into polynomial ones. For example:

• \( \ln(e^x) = x \) because the exponential function and the natural logarithm cancel each other out.

This property is crucial in simplifying and solving equations like \( e^{6x} = \frac{13}{2} \). By applying the logarithm to both sides, we are able to bring down the exponent and make it solvable through algebraic means.It's also important to recognize that the natural logarithm only applies to positive numbers. This is because you can't take the logarithm of a negative number or zero, as it is undefined in the realm of real numbers.

Exponential Functions

Exponential functions are a type of mathematical function that involves a constant base raised to a variable exponent. In the context of natural logarithms, we often use the constant \( e \) as the base, leading to expressions like \( e^x \).Exponential functions grow rapidly since each increase in the variable results in a multiplication of the base:

• Consider how \( e^x \) doubles when \( x \) increases by just 0.693 (since \( e^{0.693} \approx 2 \)).

They are powerful in modeling growth and decay processes such as population growth, radioactive decay, and interest calculations.Solving equations involving exponential functions often requires isolating the exponential term. Once isolated, applying the natural logarithm allows us to solve for the variable, as it converts the exponential growth into a manageable linear form. For instance, starting with \( e^{6x} = \frac{13}{2} \), isolating \( e^{6x} \) and then applying \( \ln \) helps in solving for \( x \). This reflects the inherent relationship between natural logarithms and exponential functions.

Inverse Operations

Inverse operations are key in solving many mathematical expressions. They allow us to "reverse" operations in equations to isolate the variable we're trying to solve. In the context of solving the exercise \( 2e^{6x} = 13 \), understanding the inverse relationship between exponentials and logarithms is vital.The principle of inverse operations states that each operation has an opposite that "undoes" it:

• Addition and subtraction are inverses.
• Multiplication and division are inverses.
• Exponential and logarithmic operations are inverses.

When dealing with \( e^{6x} \), the operation is exponential, so its inverse is taking the natural logarithm. This is how we determine the solution for \( x \):

• We apply \( \ln \) to \( e^{6x} \) to conveniently bring down the exponent and simplify the expression to \( 6x \).

This inverse relationship between exponentials and logarithms is crucial for rearranging equations and finding solutions in exponential models. Practicing this concept can greatly enhance problem-solving skills in algebra and calculus.