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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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2
Given equation is $\frac{1}{2} \log _{\sqrt{3}}\left(\frac{x+1}{x+5}\right)+\log _{9}(x+5)^{2}=1$
$\Rightarrow \quad \frac{1}{2} \cdot\left(\frac{1}{1 / 2}\right) \log _{3}\left(\frac{x+1}{x+5}\right)+\frac{1}{2} \log _{3}(x+5)^{2}=1$
$\left(\because \log _{a^{n}} b=\frac{1}{n} \log _{a} b\right)$
$\Rightarrow \quad \frac{2}{2} \log _{3}\left(\frac{x+1}{x+5}\right)+\frac{1}{2} \cdot 2 \log _{3}(x+5)=\log _{3} 3$
$\Rightarrow \quad \log _{3}\left\{\left(\frac{x+1}{x+5} \cdot(x+5)\right)\right\}=\log _{3} 3$
$\left(\because \log m+\log n=\log m n\right.$ and $\log _{n} n=1$
$\Rightarrow \quad(x+1)=3 \Rightarrow x=2$
So, only one solution is possible.
$\Rightarrow \quad \frac{1}{2} \cdot\left(\frac{1}{1 / 2}\right) \log _{3}\left(\frac{x+1}{x+5}\right)+\frac{1}{2} \log _{3}(x+5)^{2}=1$
$\left(\because \log _{a^{n}} b=\frac{1}{n} \log _{a} b\right)$
$\Rightarrow \quad \frac{2}{2} \log _{3}\left(\frac{x+1}{x+5}\right)+\frac{1}{2} \cdot 2 \log _{3}(x+5)=\log _{3} 3$
$\Rightarrow \quad \log _{3}\left\{\left(\frac{x+1}{x+5} \cdot(x+5)\right)\right\}=\log _{3} 3$
$\left(\because \log m+\log n=\log m n\right.$ and $\log _{n} n=1$
$\Rightarrow \quad(x+1)=3 \Rightarrow x=2$
So, only one solution is possible.
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