8. Sequences and Series
medium

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 $.....$

A

$1$

B

$4$

C

$3$

D

$2$

(JEE MAIN-2021)

Solution

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

Let $2^{\mathrm{x}}=\mathrm{t}$

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

$(t-5)^{2}=2 t-7$

$t^{2}-12 t+32=0$

$(t-4)(t-8)=0$

$\Rightarrow 2^{x}=4 \text { or } 2^{x}=8$

$X=2 \text { (Rejected) }$

$\text { Or } x=3$

Standard 11
Mathematics

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