Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume x > 0 and y > 0.


c. ln (x + y) = ln x + ln y

Verified step by step guidance

1. Recall the logarithm property for multiplication: \( \ln(ab) = \ln a + \ln b \), which holds for positive \(a\) and \(b\).
2. Check if the given statement \( \ln(x + y) = \ln x + \ln y \) matches this property. Notice that the left side has \(x + y\) inside the logarithm, while the right side is a sum of logarithms.
3. Since the logarithm of a sum \( \ln(x + y) \) is not equal to the sum of logarithms \( \ln x + \ln y \), this suggests the statement is generally false.
4. To confirm, consider a counterexample: choose specific positive values for \(x\) and \(y\), such as \(x = 1\) and \(y = 1\), and evaluate both sides to see if they are equal.
5. Since the values will not be equal, conclude that \( \ln(x + y) \neq \ln x + \ln y \) in general, and the statement is false.

Key Concepts

Properties of Logarithms

Logarithms have specific properties that relate sums and products. The key property is that the logarithm of a product equals the sum of the logarithms: ln(xy) = ln x + ln y. However, the logarithm of a sum, ln(x + y), does not equal the sum of logarithms, which is a common misconception.

Counterexamples in Mathematical Proof

A counterexample is a specific case that disproves a general statement. To show that ln(x + y) ≠ ln x + ln y, one can choose positive values for x and y and demonstrate that the equality does not hold, thus proving the statement false.

Domain Restrictions for Logarithmic Functions

Logarithmic functions are defined only for positive arguments. Since x > 0 and y > 0, ln x, ln y, and ln(x + y) are all defined. Understanding the domain ensures the expressions are valid and helps avoid errors when evaluating or comparing logarithmic expressions.