Question

If ${\log }_{3}2,{\log }_3\left(2^x-5\right)\mathrm{and}{\log }_3\left(2^x-\frac{7}{2}\right)$are in arithmetic progression, the value of x is

• A. 2
• B. 3
• C. 4
• D. 5

Answer 1

The correct option is B 3

Given that, ${\log }_{3}2,{\log }_3\left(2^x-5\right)\mathrm{and}{\log }_3\left(2^x-\frac{7}{2}\right)$are in arithmetic progression.
${\log }_{3}2+{\log }_3\left(2^x-\frac{7}{2}\right)=2{\log }_3\left(2^x-5\right)\Rightarrow 2\left(2^x-\frac{7}{2}\right)={\left(2^x-5\right)}^2\Rightarrow 2^{2x}-10\cdot 2^x+25-2\cdot 2^x+7=0\Rightarrow {\left(2^x\right)}^2-12\cdot 2^x+32=0\mathrm{Let}{2}^x=z.\mathrm{Then},\mathrm{we}\mathrm{get}{z}^2-12z+32=0\Rightarrow \left(z-8\right)\left(z-4\right)=0\Rightarrow z=8\mathrm{or}4\Rightarrow 2^x=8\mathrm{or}4=2^3\mathrm{or}{2}^2\Rightarrow x=3\mathrm{or}2\mathrm{But}\mathrm{for}{\log }_3\left(2^x-5\right)\mathrm{and}{\log }_3\left(2^x-\frac{7}{2}\right)\mathrm{to}\mathrm{be}\mathrm{defined}{2}^x-5>0\mathrm{and}{2}^x-\frac{7}{2}>0\Rightarrow 2^x>5\mathrm{and}{2}^x>\frac{7}{2}\Rightarrow 2^x>5\Rightarrow x\ne 2\therefore x=3$