Question

If ${\log }_{3}2,{\log }_3(2^x-5)$ and ${\log }_3(2^x-\frac{7}{2})$ are in A.P., then $x$ is equal to

• A. $1,\frac{1}{2}$
• B. $1,\frac{5}{2}$
• C. $1,\frac{3}{2}$
• D. None of these

Answer 1

Answer: (D)

None of these

${\log }_{3}2,{\log }_3(2^x-5)$ and ${\log }_3(2^x-\frac{7}{2})$ are in A.P

$\Rightarrow {\log }_3(2^x-5)-{\log }_{3}2={\log }_3(2^x-\frac{7}{2})-{\log }_3(2^x-5)$

$\Rightarrow {\log }_3\left(\frac{2^x-5}{2}\right)={\log }_3\left(\frac{2^x-\frac{7}{2}}{2^x-5}\right)$

$\Rightarrow \frac{2^x-5}{2}=\frac{2^x-\frac{7}{2}}{2^x-5}$

$\Rightarrow {(2^x-5)}^2=2(2^x-\frac{7}{2})$

$\Rightarrow 2^{2x}+25-10\times 2^x=2\times 2^x-7$

$\Rightarrow 2^{2x}-12\left(2^x\right)+32=0$

Let $\alpha =2^x\Rightarrow {\alpha }^2-12\alpha +32=0$

$\Rightarrow {\alpha }^2-4\alpha -8\alpha +32=0$

$\Rightarrow \alpha (\alpha -4)-8(\alpha -4)=0$

$\Rightarrow \alpha =8,4$

$\Rightarrow 2^x=2^3,2^2$

$\therefore x=3,2$

When $x=3$,$2^x-5>0$.Hence $logx$ is defined when $x>0$

When $x=2$,$2^x-5<0$.where the $\log$ is not defined.

Hence $x=3$ is the only solution.