Question

If $f(x) = \log_{(x-1)^2}(x^2-3x+2)$, $x \in \mathbb{R}-[1,2]$ and $x\neq0$, then $f'(3)=$

• A. 1
• B. 0
• C. $\log_e 4$
• D. $\log_4 e$

Hint:

When dealing with logarithmic functions with a variable in the base, the first step should always be to use the change of base formula to convert to a standard base like $e$ or 10. This makes differentiation much more straightforward.

Answer 1

The Correct Option is D

Step 1: Simplify the function The base of the logarithm is \( x^2-2x+1 = (x-1)^2 \). The argument is \( x^2-3x+2 = (x-1)(x-2) \). Using the change of base property \( \log_a b = \frac{\ln b}{\ln a} \): \[ f(x) = \frac{\ln((x-1)(x-2))}{\ln((x-1)^2)} = \frac{\ln(x-1) + \ln(x-2)}{2\ln(x-1)} \] \[ f(x) = \frac{1}{2} + \frac{1}{2} \frac{\ln(x-2)}{\ln(x-1)} \]
Step 2: Differentiate the function Use the quotient rule for the second term: \[ f'(x) = \frac{1}{2} \left[ \frac{\ln(x-1) \cdot \frac{1}{x-2} - \ln(x-2) \cdot \frac{1}{x-1}}{(\ln(x-1))^2} \right] \]
Step 3: Calculate \( f'(3) \) Substitute \( x = 3 \): Terms involved: \( x-1 = 2 \), \( \ln(x-1) = \ln 2 \). \( x-2 = 1 \), \( \ln(x-2) = \ln 1 = 0 \). Substitute these values: \[ f'(3) = \frac{1}{2} \left[ \frac{(\ln 2) \cdot \frac{1}{1} - 0 \cdot \frac{1}{2}}{(\ln 2)^2} \right] \] \[ f'(3) = \frac{1}{2} \frac{\ln 2}{(\ln 2)^2} = \frac{1}{2 \ln 2} = \frac{1}{\ln 4} \] Since \( \frac{1}{\ln 4} = \log_4 e \): \[ f'(3) = \log_4 e \]