If log _{3} 2, log _{3}left(2^{x}-5right), lo | vedclass

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Question

$2 \log _{3}\left(2^{x}-5\right)=\log _{3} 2+\log _{3}\left(2^{x}-\frac{7}{2}\right)$

Let $2^{\mathrm{x}}=\mathrm{t}$

$\log _{3}(t-5)^{2}=\log _

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$2 \log _{3}\left(2^{x}-5ight)=\log _{3} 2+\log _{3}\left(2^{x}-\frac{7}{2}ight)$

Let $2^{\mathrm{x}}=\mathrm{t}$

$\log _{3}(t-5)^{2}=\log _{3} 2\left(t-\frac{7}{2}ight)$

$(t-5)^{2}=2 t-7$

$t^{2}-12 t+32=0$

$(t-4)(t-8)=0$

$\Rightarrow 2^{x}=4 	ext { or } 2^{x}=8$

$X=2 	ext { (Rejected) }$

$	ext { Or } x=3$

If $\log _{3} 2, \log _{3}\left(2^{x}-5\right), \log _{3}\left(2^{x}-\frac{7}{2}\right)$ are in an arithmetic progression, then the value of $x$ is equal to $.....$

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