Question

In Exercises, determine whether the statement is true or false given that \(f(x)=\ln x .\) If it is false, explain why or give an example that shows it is false. $$ \sqrt{f(x)}=\frac{1}{2} f(x) $$

Short answer

The statement \( \sqrt{f(x)} = \frac{1}{2} f(x) \) is false for the function \( f(x) = \ln x \). A counterexample is \( x = 0.5 \), for which we have \( \sqrt{\ln 0.5} > \frac{1}{2} \ln 0.5 \)

Step-by-step solution

Substitute the given function

Substitute \( f(x) = \ln x \) into the equation \( \sqrt{f(x)} = \frac{1}{2} f(x) \). This gives \( \sqrt{\ln x} = \frac{1}{2} \ln x \).

Analyze the equation

Looking carefully at the equation, it's clear that this will be true if and only if \( \ln x \) is always non-negative. That's because a positive number when divided by 2 is greater than its square root, while a negative number when divided by 2 is less than its square root.

Check the domain

Check the domain of the function \( f(x) = \ln x \) which is \( x > 0 \). Since \( \ln x \) is negative for \( 0 < x < 1 \), it follows that there exist numbers in the domain of the function \( f(x) = \ln x \) that make \( \sqrt{f(x)} > \frac{1}{2} f(x) \). For example, try substituting \( x = 0.5 \) into both sides of the equation.

Counterexample

For \( x = 0.5 \), we have \( \sqrt{\ln 0.5} = 0.94 \) and \( \frac{1}{2} \ln 0.5 = -0.35 \). Therefore, the given statement is false, and \( x = 0.5 \) is a counterexample.

Key concepts

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions. They have the form f(x) = log_b x, where b is the base of the logarithm, and x is the argument. Logarithms are used to solve for the exponent in an equation where the exponent's base and its result are known.

For example, if we have b^y = x, then the logarithm base b of x is y, written as y = log_b x. A special kind of logarithm is the natural logarithm, denoted as ln, which has the base e (Euler's number, approximately 2.71828).

Understanding logarithmic functions requires familiarity with their properties, such as:

• Product rule: log_b(xy) = log_b x + log_b y
• Quotient rule: log_b(x/y) = log_b x - log_b y
• Power rule: log_b(x^r) = r * log_b x
• Change of base rule: log_b x = log_k x / log_k b, for any positive k (≠1)

Domain of Logarithm

The domain of a logarithmic function is a critical concept as it concerns the set of inputs for which the function is defined. For the natural logarithm function, or any logarithmic function f(x) = log_b x, the domain consists of all positive real numbers. In other words, only positive numbers can be placed inside a logarithm because you cannot take the logarithm of a negative number, nor of zero.

This condition is rooted in the function's inverse relationship with exponents. Since an exponent can only result in a positive outcome when the base is positive (ignoring complex numbers for the scope of this discussion), the input to its inverse logarithm must also be positive.

For instance, considering the function f(x) = ln x, the logarithm only takes positive values of x to yield real numbers. Hence, the domain for this function is x > 0. Understanding a function's domain is vital in solving equations and inequalities involving logarithms.

Square Roots of Logarithms

Exploring the square roots of logarithms entails examining the outcome when applying the square root function to a logarithmic function. It's essential to remember that taking the square root of a logarithm is not the same as multiplying it by one-half.

An immediate example comes from the equation \( \sqrt{f(x)} = \frac{1}{2} f(x) \), where f(x) = ln x. Without understanding the domain and properties of logarithms, one might easily mistake this equation for being true under all circumstances. However, as the step-by-step solution indicates, this statement is false because, within the domain of f(x), \( \ln x \) can yield negative values especially for x between 0 and 1. For these values, the square root is undefined.

Therefore, while exploring square roots of logarithms, it is crucial to consider the domain of the logarithm involved and to validate the original equation's accuracy for all possible values within that domain. As demonstrated in the counterexample, verifying the truth of expressions involving square roots of logarithms must involve a careful examination, acknowledging that the relationship is not a straightforward proportionality.