MCQ
If $\log _{10} 2, \log _{10} (2^x + 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-
- A$x = 0$
- B$x = 1$
- C$x = \log _{10} 2$
- ✓$x = \frac{1}{2} \log _2 5$
$\Rightarrow 2 \log _{10}\left(2^{x}-1\right)=\log _{10} 2+\log _{10}\left(2^{x}+3\right)$
$\Rightarrow \log _{10}\left(2^{x}-1\right)^{2}=\log _{10} 2\left(2^{x}+3\right)$
$\Rightarrow \log _{10}\left(2^{2 x}+1-2^{x+1}\right)=\log _{10}\left(2^{x+1}+6\right)$
$\Rightarrow 2^{2 x}+1-2^{x+1}=2^{x+1}+6$
$\Rightarrow 2^{2 x}-2^{x+2}-5=0$
Take, $2^{x}=y$
$\Rightarrow y^{2}-4 y-5=0$
$\Rightarrow(y-5)(y+1)=0$
$\Rightarrow y=5,-1$
$\because y>0 \Rightarrow y=5$
$\Rightarrow 2^{x}=5$
$\Rightarrow x=\log _{2} 5$
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where $\tan ^{-1} x$ takes only principal values, then the value of $\left(\log _8|1+\alpha|-\frac{3 \pi}{4}\right)$ is