Question

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

Answer 1

Considering the question to be:

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

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

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

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

Let $y=2^x$

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

$\Rightarrow 2y-7=y^2+25-10y$

$y^2-12y+32=0$

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

$y=4,8$

$i.e,2^x=4,8$

$\Rightarrow x=2,3$

$\therefore x=3$, because when $x=2$ 2nd term is undefined