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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