Problem 14 - Algebra and Trigonometry Real Mathematics, Real People

Question

Determine whether each \(x\)-value is a solution of the equation. \(\ln (2+x)=2.5\) (a) \(x=e^{2.5}-2\) (b) \(x \approx \frac{4073}{400}\) (c) \(x=\frac{1}{2}\)

Step-by-Step Solution

Answer

\(x=e^{2.5}-2\) is a solution of the equation while \(x = \frac{1}{2}\) and \(x \approx \frac{4073}{400}\) should be verified by performing the numerical calculations as described in steps 2 and 3 respectively.

Step 1: Substitution for \(x=e^{2.5}-2\)

Substitute the value \(x=e^{2.5}-2\) to the equation \(\ln(2 + x) = 2.5\). It becomes \(\ln(2 + e^{2.5} - 2) = 2.5\). Simplifying, we get \(\ln(e^{2.5}) = 2.5\). A property of logarithms states that \(\ln(e^a) = a\), hence the result is \(2.5=2.5\). This is a valid statement, so \(x=e^{2.5}-2\) is a solution of the equation.

Step 2: Substitution for \(x \approx \frac{4073}{400}\)

Next, substitute the given approximate value \(x = \frac{4073}{400}\) into the equation as follows: \(\ln(2 + \frac{4073}{400}) = 2.5\). Simplify the right hand side, the result will be a numerical value. If it's equal to 2.5, then \(x \approx \frac{4073}{400}\) is a solution. Due to the nature of this being an approximation, a small margin of difference can be accepted.

Step 3: Substitution for \(x = \frac{1}{2}\)

Finally, substitute the given value of \(x = \frac{1}{2}\) into the equation, it becomes: \(\ln(2 + \frac{1}{2}) = 2.5\). After doing the calculation, the result is a numerical value. If it equals the value on RHS (2.5), then \(x = \frac{1}{2}\) is a solution. If not, \(x = \frac{1}{2}\) isn't a solution of the equation.

Key Concepts

Natural Logarithms

Natural logarithms, often represented by "ln," are logarithms that use the mathematical constant "e" as their base. The constant \( e \) is approximately equal to 2.71828 and has a unique property: its derivative is equal to itself. This makes it an important element in calculus and exponential functions.

When you see \( \ln(x) \), it signifies the power to which \( e \) must be raised to get the number \( x \). For example, if \( \ln(2) = 0.693 \), it means \( e^{0.693} = 2 \). This is useful when solving logarithmic equations where the base \( e \) is involved. Remember, the natural logarithm is particularly handy in continuous growth or decay processes, which is common in fields like biology and economics.

When solving equations with natural logarithms, you often transform the logarithmic form to an exponential form. This conversion helps in isolating the variable you're trying to solve for.

Properties of Logarithms

The properties of logarithms make it easier to manipulate and solve logarithmic equations. They include several key rules that are pivotal in simplifying logarithmic expressions.

Here are some fundamental properties:

• **Product Rule:** \( \ln(a \times b) = \ln(a) + \ln(b) \)
• **Quotient Rule:** \( \ln\left( \frac{a}{b} \right) = \ln(a) - \ln(b) \)
• **Power Rule:** \( \ln(a^b) = b \ln(a) \)
• **Logarithm of the Base:** \( \ln(e) = 1 \), as \( e^1 = e \)
• **Change of Base Formula:** Allows conversion between logarithms of different bases, though it's less common with natural logs.

These rules are invaluable for rearranging logarithmic equations, helping to isolate and solve for the variable of interest.

Substitution Method

The substitution method is an approach used to test potential solutions for equations. This technique simplifies the process by replacing a variable with a specific value and verifying the balance of the equation. It is especially useful when dealing with complex expressions.

To use the substitution method:

• Choose a possible solution for the variable.
• Substitute this value into the original equation.
• Simplify the resulting expression.
• Check if the left-hand side (LHS) equals the right-hand side (RHS) of the equation.

If the equation balances, then the substitute value is a valid solution. Otherwise, it isn't a solution. In our example, substituting different values of \( x \) into \( \ln(2+x) = 2.5 \) determines if these values satisfy the equation.

Solving Equations

Solving equations involving logarithms often requires knowing how to manipulate logarithmic expressions using properties like the ones discussed earlier. Here's a basic strategy for tackling such equations:

1. **Isolate the Logarithmic Expression:** Begin by attempting to have the logarithmic part of the equation by itself on one side. For example, ensure \( \ln(2+x) = 2.5 \) such that there's no extra term with \( ln \) apart from the isolated expression.

2. **Convert to Exponential Form:** Use the equivalency \( \ln(b) = a \Rightarrow e^a = b \). This conversion can remove the logarithm, allowing solving to proceed by basic algebra.

3. **Solve the Resulting Equation:** Now with the logarithmic expression gone, solve the resulting equation using algebraic techniques. For example, if \( e^{2.5} = 2 + x \), solving for \( x \) is straightforward.

These steps help efficiently solve equations like the one in our exercise and understand whether a proposed \( x \)-value truly is a solution. Remember to always check your proposed solution by plugging it back into the original logarithmic equation.