The number of solutions of the equation log _ | vedclass

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

Question

(A) Given equation: $\log _{4}(x-1)=\log _{2}(x-3)$Using the property $\log _{a^n}(b) = \frac{1}{n} \log _{a}(b)$,we have:$\frac{1}{2} \log _{2}(x-1)=

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) Given equation: $\log _{4}(x-1)=\log _{2}(x-3)$Using the property $\log _{a^n}(b) = \frac{1}{n} \log _{a}(b)$,we have:$\frac{1}{2} \log _{2}(x-1)=\log _{2}(x-3)$$\log _{2}(x-1)^{1/2}=\log _{2}(x-3)$Equating the arguments:$(x-1)^{1/2} = x-3$Squaring both sides:$x-1 = (x-3)^2$$x-1 = x^2 - 6x + 9$$x^2 - 7x + 10 = 0$Factoring the quadratic equation:$(x-2)(x-5) = 0$$x = 2$ or $x = 5$Check the domain of the original equation:For $\log _{2}(x-3)$ to be defined,$x-3 > 0$,so $x > 3$.For $\log _{4}(x-1)$ to be defined,$x-1 > 0$,so $x > 1$.Combining these,the domain is $x > 3$.Since $x=2$ is not in the domain $(2 Thus,only $x=5$ is a valid solution.Therefore,the number of solutions is $1$.

The number of solutions of the equation $\log _{4}(x-1)=\log _{2}(x-3)$ is

Similar Questions

$\log_{7} \log_{7} \sqrt{7 \sqrt{7 \sqrt{7}}} = ?$

If $\log _{2}\left(9^{x-1}+7\right)-\log _{2}\left(3^{x-1}+1\right)=2$,then $x$ values are

The number $\log_{20} 3$ lies in

$\operatorname{coth}^{-1} 3 + \tanh^{-1} \frac{1}{3} - \operatorname{cosech}^{-1}(-\sqrt{3}) = $

Let $\phi (x) = x + 2^{\log_x 3} - 3^{\log_x 2}$. Then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module