Problem 23 - College Algebra with Corequisite Support

Question

For the following exercises, use logarithms to solve. \(10 e^{8 x+3}+2=8\)

Step-by-Step Solution

Answer

\( x = \frac{\ln(0.6) - 3}{8} \)

Step 1: Isolate the Exponential Term

First, subtract 2 from both sides of the equation to isolate the term with the exponential. This gives us:\[ 10e^{8x+3} = 6 \]

Step 2: Divide to Further Isolate the Exponential Term

Next, divide both sides of the equation by 10 to isolate the exponential term:\[ e^{8x+3} = 0.6 \]

Step 3: Apply the Natural Logarithm

Take the natural logarithm (ln) of both sides to eliminate the exponential. This step uses the property that \( \ln(e^y) = y \):\[ \ln(e^{8x+3}) = \ln(0.6) \]Therefore, this simplifies to:\[ 8x + 3 = \ln(0.6) \]

Step 4: Solve for x

Now, solve for \( x \) by first subtracting 3 from both sides:\[ 8x = \ln(0.6) - 3 \]Then divide by 8:\[ x = \frac{\ln(0.6) - 3}{8} \]

Key Concepts

Exponential Functions

Exponential functions are a fundamental concept in mathematics dealing with equations where a constant base is raised to a variable exponent. In the problem given, the exponential term is represented by \( e^{8x+3} \). The letter \( e \) stands for Euler's number, a special mathematical constant approximately equal to 2.71828. Exponential functions have the form \( a^x \), where \( a \) is a positive constant not equal to one.

• They are crucial in modeling growth and decay scenarios like population growth, radioactive decay, and continuously compounded interest.
• Key features include the rate of change, which is proportional to the current value, leading to rapid rises or declines.

When solving equations with exponential terms, the main goal is to isolate the exponent on one side of the equation, making it easier to use logarithmic functions to find the value of the variable.

Natural Logarithm

The natural logarithm, denoted as \( \ln \), is the logarithm to the base \( e \). It is the inverse function of the exponential \( e^x \). Using natural logarithms is particularly convenient when you have exponential equations involving \( e \), as they simplify back to linear terms.

• The property \( \ln(e^x) = x \) is especially helpful as it effectively "undoes" the exponentiation by \( e \).
• Natural logarithms help us convert multiplicative relationships into additive ones, which are easier to manipulate and solve.

In the provided step-by-step solution, applying \( \ln \) to both sides of the equation \( e^{8x+3} = 0.6 \) was a pivotal move that reduced the problem from an exponential form into a more straightforward algebraic expression: \( 8x + 3 = \ln(0.6) \). This transformation is key in many calculus and algebra problems.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate the desired variable. This method uses basic arithmetic operations: addition, subtraction, multiplication, and division. The process simplifies complex equations into more manageable forms and equations.

• Start by isolating terms involving the variable using inverse operations.
• Solve multi-step equations by systematically applying algebraic rules to both sides.

In the exercise, after applying the natural logarithm, algebraic manipulation was used to reach the solution for \( x \). This included subtracting 3 from both sides of \( 8x + 3 = \ln(0.6) \) to further isolate \( 8x \), and finally, dividing by 8 to solve for \( x \), this leads to the final expression \( x = \frac{\ln(0.6) - 3}{8} \). This step-by-step organization allows a systematic approach to solving equations efficiently, ensuring accuracy and clarity.