If log _{10} 2, log _{10} (2^x + 1), log _{10 | vedclass

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Question

If $a, b, c$ are in $AP$ then $2 b=a+c$

$\Rightarrow 2 \log _{10}\left(2^{x}-1\right)=\log _{10} 2+\log _{10}\left(2^{x}+3\right)$

$\Rightarrow

Vedclass · Accepted Answer

$x = \frac{1}{2} \log _2 5$

Vedclass · Answer

If $a, b, c$ are in $AP$ then $2 b=a+c$

$\Rightarrow 2 \log _{10}\left(2^{x}-1\right)=\log _{10} 2+\log _{10}\left(2^{x}+3\right)$

$\Rightarrow \log _{10}\left(2^{x}-1\right)^{2}=\log _{10} 2\left(2^{x}+3\right)$

$\Rightarrow \log _{10}\left(2^{2 x}+1-2^{x+1}\right)=\log _{10}\left(2^{x+1}+6\right)$

$\Rightarrow 2^{2 x}+1-2^{x+1}=2^{x+1}+6$

$\Rightarrow 2^{2 x}-2^{x+2}-5=0$

Take, $2^{x}=y$

$\Rightarrow y^{2}-4 y-5=0$

$\Rightarrow(y-5)(y+1)=0$

$\Rightarrow y=5,-1$

$\because y>0 \Rightarrow y=5$

$\Rightarrow 2^{x}=5$

$\Rightarrow x=\log _{2} 5$

If $\log _{10} 2, \log _{10} (2^x + 1), \log _{10} (2^x + 3)$ are in $A.P.,$ then :-

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