The number of solutions of the equation frac{ | vedclass

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(B) Given equation is $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$. For the logarithm to be defined,we must have $\

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$1$

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(B) Given equation is $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}ight)+\log _{9}(x+5)^{2}=1$. For the logarithm to be defined,we must have $\frac{x+1}{x+5} > 0$ and $x+5 
eq 0$. This implies $x \in (-\infty, -5) \cup (-1, \infty)$. Using the property $\log_{a^n} b = \frac{1}{n} \log_a b$,we have $\log_{\sqrt{3}} y = \log_{3^{1/2}} y = 2 \log_3 y$ and $\log_9 y = \log_{3^2} y = \frac{1}{2} \log_3 y$. The equation becomes $\frac{1}{2} \cdot 2 \log_3 \left(\frac{x+1}{x+5}ight) + \frac{1}{2} \cdot 2 \log_3 |x+5| = 1$. $\log_3 \left(\frac{x+1}{x+5}ight) + \log_3 |x+5| = 1$. Since $x+5$ must be positive for the domain,$|x+5| = x+5$. $\log_3 \left(\frac{x+1}{x+5} \cdot (x+5)ight) = 1$. $\log_3 (x+1) = 1$. $x+1 = 3^1 = 3$,so $x = 2$. Since $x=2$ satisfies the domain condition,there is only $1$ solution.

The number of solutions of the equation $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$ is

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