Problem 5 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

For what value of \(x\) function be identical \(f(x)=\log (x-1)-\log (x-2)\) and \(g(x)=\) \(\log \left(\frac{x-1}{x-2}\right)\) (a) \((1, \infty)\) (b) \((2, \infty)\) (c) \((3, \infty)\) (d) None of these

Step-by-Step Solution

Answer

(b) (2, ∞)

Step 1: Understand the Problem

The given problem asks for the value of \(x\) where two functions, \(f(x) = \log (x-1) - \log (x-2)\) and \(g(x) = \log \left(\frac{x-1}{x-2}\right)\), are identical. This means they should have the same domain, range, and algebraic form.

Step 2: Recognize the Logarithmic Property

Using the property of logarithms, \(\log a - \log b = \log \left( \frac{a}{b} \right)\), we can simplify \(f(x)\). Therefore, \(f(x) = \log (x-1) - \log (x-2) = \log \left(\frac{x-1}{x-2}\right)\), which is identical to \(g(x)\).

Step 3: Determine the Domain

The logarithm function \(\log a\) is defined only for \(a > 0\). So for \(f(x)\):1. \(x-1 > 0 \Rightarrow x > 1\).2. \(x-2 > 0 \Rightarrow x > 2\).Taking both conditions into account, the domain for \(f(x)\) is \((2, \infty)\).

Step 4: Compare with Options

From Step 3, the domain where both functions have the same form and are defined is \((2, \infty)\). Comparing this with the given options, option (b) \((2, \infty)\) is correct.

Key Concepts

Understanding Logarithmic Properties

Logarithmic functions have unique properties that can simplify complex expressions. One crucial property is the difference law:

• \(\log a - \log b = \log \left( \frac{a}{b} \right)\)

This property is used to convert a subtraction of logarithms into a single logarithm. Such transformations help in solving, simplifying, and analyzing logarithmic equations.
In our exercise, we applied this property to show that the function \(f(x) = \log (x-1) - \log (x-2)\) is identical to \(g(x) = \log \left(\frac{x-1}{x-2}\right)\). This crucial step relied on our understanding of how logarithmic expressions can be manipulated using logarithmic properties.
This illustrates the power and convenience of logarithmic properties in making two seemingly different expressions equivalent.

Exploring Function Domains

The domain of a function refers to all possible input values that will produce a valid output. For logarithmic functions, this means ensuring the argument of the logarithm is positive, since the logarithm of a non-positive number is undefined.
In our problem, we analyzed the domain of \(f(x) = \log (x-1) - \log (x-2)\) by applying the necessary condition:

• \(x-1 > 0 \Rightarrow x > 1\)
• \(x-2 > 0 \Rightarrow x > 2\)

When we consider these inequalities together, the stricter condition \(x > 2\) dictates the domain of our function. Therefore, the domain is \((2, \infty)\).
Understanding domains helps in identifying valid inputs for functions, and in this context, ensures the expressions within our logarithms remain valid.

Solving Logarithmic Equations

Logarithmic equations involve the unknown variable within the logarithm itself. Solving these equations often requires rearranging using logarithmic properties or applying restrictions based on domains.
For example, in the problem presented, the challenge was to determine when two logarithmic expressions, \(f(x)\) and \(g(x)\), are identical. By employing the property \(\log a - \log b = \log \left( \frac{a}{b} \right)\), we simplified \(f(x)\) to match it algebraically and structurally with \(g(x)\).
Besides algebraic manipulation, solving such equations requires us to ensure that all assumptions regarding the domain are satisfied. This involves verifying that the values of \(x\) keep the arguments of all logarithms positive. Hence, the solution \((2, \infty)\) was chosen as it satisfies the condition for both functions being defined.