If 2 log (x+1)-log (x^{2}-1)=log 2, then x= | vedclass

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

Question

(A) Given the equation: $2 \log (x+1)-\log (x^{2}-1)=\log 2$. Using the property $n \log a = \log (a^n)$, we get $\log (x+1)^2 - \log (x^2-1) = \log 2

Vedclass · Accepted Answer

only $3$

Vedclass · Answer

(A) Given the equation: $2 \log (x+1)-\log (x^{2}-1)=\log 2$. Using the property $n \log a = \log (a^n)$, we get $\log (x+1)^2 - \log (x^2-1) = \log 2$. Using the property $\log a - \log b = \log (a/b)$, we get $\log \left( \frac{(x+1)^2}{x^2-1} \right) = \log 2$. Since $x^2-1 = (x-1)(x+1)$, the equation becomes $\log \left( \frac{(x+1)^2}{(x-1)(x+1)} \right) = \log 2$. Simplifying the fraction, we have $\log \left( \frac{x+1}{x-1} \right) = \log 2$. Therefore, $\frac{x+1}{x-1} = 2$. $x+1 = 2(x-1) \Rightarrow x+1 = 2x-2$. $x = 3$. For the logarithmic terms to be defined, we require $x+1 > 0$ and $x^2-1 > 0$, which implies $x > 1$. Thus, $x = 3$ is the only valid solution.

If $2 \log (x+1)-\log (x^{2}-1)=\log 2$, then $x=$

Similar Questions

The equation $x^{(\log _{3} x)^{2}-\frac{9}{2} \log _{3} x+5}=3 \sqrt{3}$ has

The value of $\log _4 18$ is

The sum of the roots of the equation $x+1-2 \log _{2}\left(3+2^{x}\right)+2 \log _{4}\left(10-2^{-x}\right)=0$ is:

If $\log _k x \cdot \log _5 k = \log _x 5$,where $k \ne 1$ and $k > 0$,then $x$ is equal to:

If $3^x = 4^{x-1}$,then $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module