If f(x) = log_{(x^2-2x+1)}(x^2-3x+2), x in R | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given $f(x) = \log_{(x^2-2x+1)}(x^2-3x+2)$.We can rewrite the base and the argument as:$x^2-2x+1 = (x-1)^2$$x^2-3x+2 = (x-1)(x-2)$So, $f(x) = \log

Vedclass · Accepted Answer

$\log_4 e$

Vedclass · Answer

(D) Given $f(x) = \log_{(x^2-2x+1)}(x^2-3x+2)$.We can rewrite the base and the argument as:$x^2-2x+1 = (x-1)^2$$x^2-3x+2 = (x-1)(x-2)$So, $f(x) = \log_{(x-1)^2} ((x-1)(x-2))$.Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$, we get:$f(x) = \frac{1}{2} \log_{(x-1)} ((x-1)(x-2)) = \frac{1}{2} [\log_{(x-1)} (x-1) + \log_{(x-1)} (x-2)]$$f(x) = \frac{1}{2} [1 + \log_{(x-1)} (x-2)] = \frac{1}{2} [1 + \frac{\ln(x-2)}{\ln(x-1)}]$.Now, differentiate with respect to $x$:$f'(x) = \frac{1}{2} \left[ \frac{\ln(x-1) \cdot \frac{1}{x-2} - \ln(x-2) \cdot \frac{1}{x-1}}{(\ln(x-1))^2} \right]$.Substitute $x = 3$:$f'(3) = \frac{1}{2} \left[ \frac{\ln(2) \cdot \frac{1}{1} - \ln(1) \cdot \frac{1}{2}}{(\ln(2))^2} \right] = \frac{1}{2} \left[ \frac{\ln(2)}{(\ln(2))^2} \right] = \frac{1}{2 \ln(2)} = \frac{1}{\ln(4)} = \log_4 e$.

If $f(x) = \log_{(x^2-2x+1)}(x^2-3x+2)$, $x \in R - \{1, 2\}$ and $x \neq 0$, then $f'(3) =$

Similar Questions

If $f(x) = e^{g(x)}$ and $g(x) = \int_{2}^{x} \frac{t}{1 + t^4} \, dt$,then $f'(2)$ has the value equal to:

Differentiate the following with respect to $x$: $\log(\cos e^{x})$

If $f(x)=\log _e\left(e^{2 x}\left(\frac{3 x+5}{5-3 x}\right)^{\frac{2}{3}}\right)$, $x \neq \frac{-5}{3}, \frac{5}{3}$, then the value of $\frac{d f}{d x}$ at $x=1$ is

$\frac{d}{dx} \left[ \log \left\{ e^x \left( \frac{x - 2}{x + 2} \right)^{3/4} \right\} \right]$ is equal to

If $f(1) = 3$ and $f'(1) = 2$,then find the value of $\frac{d}{dx} \{ \log f(e^x + 2x) \}$ at $x = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module