Problem 10 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether each \(x\)-value is a solution of the equation. \(4 e^{x-1}=60\) (a) \(x=1+\ln 15\) (b) \(x \approx 3.7081\) (c) \(x=\ln 16\)

Step-by-Step Solution

Answer

The only value that is a solution to the equation is \(x = 1+ \ln15\). Both \(x \approx 3.7081\) and \(x = \ln16\) do not satisfy the equation.

Step 1: Checking x = 1+ ln15

Substitute \(x=1+\ln15\) into the equation: \(4e^{1+\ln15 - 1}= 4e^{\ln15}= 4*15 =60\). As you can see, it yields the left side of the equation equal to the right side, proving that \(x=1+\ln15\) is indeed a solution to the equation.

Step 2: Checking x ≈ 3.7081

Substitute \(x \approx 3.7081\) into the equation: \(4e^{3.7081 - 1}\). By calculating this expression, you will find that it doesn't equal to 60, meaning that \(x \approx 3.7081\) is not a solution to the equation.

Step 3: Checking x = ln16

Substitute \(x=\ln16\) into the equation: \(4e^{\ln16 - 1}\). After calculating the left side, you will find that it doesn't equal to 60, hence \(x=\ln16\) is not a solution to the equation.

Key Concepts

Solving Equations

When faced with equations like \(4e^{x-1} = 60\), the goal is to find values of \(x\) that make the equation true. In other words, we need the left side to equal the right side when substituted with our \(x\)-value. This process involves:

• Substituting the \(x\)-value into the equation.
• Simplifying the expression using the properties of exponents and logarithms.
• Ensuring that both sides of the equation remain equal.

Solving equations step-by-step can be tedious, but it lets us verify any possible solution. Substituting suggested \(x\)-values individually is crucial, as each calculation checks whether the specific \(x\) satisfies the equation. This ensures accuracy and a greater understanding of the solution.

Natural Logarithms

Natural logarithms are mathematical operations associated with the base of Euler's number \(e\). They can be particularly handy when solving exponential equations, such as when isolating \(x\) in the equation \(4e^{x-1} = 60\). The natural logarithm, denoted as \(\ln\), helps to 'undo' the exponential function, revealing \(x\):

• Applying \(\ln\) to both sides of an equation allows us to bring down the exponent, simplifying the expression.
• Remember that \(\ln(e^x) = x\), which makes finding \(x\) quite straightforward.

In this exercise, the natural logarithm played a role when checking \(x = 1 + \ln 15\), where it successfully broke down the exponential component, confirming the correctness of the expression's simplification.

Exponential Functions

Exponential functions are characterized by an equation in which a constant base is raised to a variable exponent, like \(e^x\). These functions often describe growth or decay processes, with the base \(e\) being a fundamental mathematical constant.The prime focus of exponential equations is getting both sides balanced. This often involves:

• Recognizing and using the property that exponential functions with the same base can be compared directly.
• Using logarithms to isolate the variable in an exponential form.

In the given problem, the equation \(4e^{x-1} = 60\) required evaluating various potential solutions by adjusting exponents. Understanding how exponents contribute to solving these equations can deepen comprehension of how changes in \(x\) affect the outcome of exponential expressions.