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