Problem 12 - College Algebra with Corequisite Support

Question

For the following exercises, use logarithms to solve. \(2 e^{6 x}=13\)

Step-by-Step Solution

Answer

\(x = \frac{\ln\left(\frac{13}{2}\right)}{6}\).

Step 1: Isolate the Exponential Term

To start, isolate the exponential expression on one side of the equation. The given equation is \(2 e^{6x} = 13\). Divide both sides by 2 to get \(e^{6x} = \frac{13}{2}\).

Step 2: Apply the Natural Logarithm

Apply the natural logarithm (\(\ln\)) to both sides of the equation to eliminate the exponential expression. This gives us \(\ln(e^{6x}) = \ln\left(\frac{13}{2}\right)\).

Step 3: Simplify the Logarithmic Expression

Utilize the logarithmic identity \(\ln(a^b) = b\cdot \ln(a)\). Applying this identity, \(\ln(e^{6x})\) simplifies to \(6x\cdot \ln(e)\). Since \(\ln(e) = 1\), the equation becomes \(6x = \ln\left(\frac{13}{2}\right)\).

Step 4: Solve for x

Finally, solve for \(x\) by dividing both sides by 6. Thus, \(x = \frac{\ln\left(\frac{13}{2}\right)}{6}\).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. They are crucial in various fields, including science, finance, and statistics.
One common base used in exponential functions is the natural number "e," approximately equal to 2.71828. An expression like \(e^{6x}\) shows how an exponential function grows or decays with respect to the variable \(x\).

• This growth or decay happens rapidly due to the constant multiplier effect of the base \(e\).
• Exponential functions are often used to model real-world scenarios such as population growth, radioactive decay, or compound interest.

Understanding how to manipulate and solve equations with exponential functions, like isolating \(e^{6x}\) in our example, is a fundamental skill in algebra, enabling us to better understand and predict patterns in natural phenomena.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\). This mathematical operation is essential for solving equations involving exponential expressions.
The natural logarithm allows us to reverse the exponential function, transforming a multiplication process into an addition one, thanks to the logarithmic identities.

• For instance, If you have the expression \(\ln(a^b)\), this simplifies to \(b\cdot \ln(a)\). This property is used extensively in algebra to simplify and solve equations.
• One of the most important features of the natural logarithm is that \(\ln(e) = 1\), which simplifies calculations, as seen where \(6x \cdot \ln(e)\) becomes \(6x\).

The natural logarithm is central when working with continuous growth or decay, allowing easy manipulation and solution of equations, such as the conversion of \(e^{6x} = \frac{13}{2}\) into a solvable form.

Solving Equations

Solving equations that include exponential and logarithmic components involves a structured set of steps to isolate and simplify terms.
The process typically includes:

• Isolating the exponential term: Start by ensuring the exponential expression stands alone on one side of the equation, as shown with \(2 e^{6x} = 13\), which simplifies to \(e^{6x} = \frac{13}{2}\).
• Applying logarithms: Incorporate the natural logarithm to eliminate the exponential. This allows you to apply algebraic manipulation using logarithmic identities.
• Simplifying the expression: Utilize properties of logarithms, for instance, turning \(\ln(e^{6x})\) into \(6x\), which simplifies the calculation process.
• Solving for the variable: Conclude by performing basic algebraic operations, such as division, to solve for the variable \(x\) as demonstrated by \(x = \frac{\ln\left(\frac{13}{2}\right)}{6}\).

By following these structured steps, solving equations with exponential functions becomes a manageable task, clearing pathways to understanding more complex algebraic concepts.