Question

If ${\log }_{3}2,{\log }_3(2^x-5),{\log }_3(2^x-\frac{7}{2})$ are in arithmetic progression , determine the value of $x$.

Answer 1

${\log }_{3}2,{\log }_3(2^x-5),{\log }_3(2^x-\frac{7}{2})$ are in arithmetic progression.
$\Rightarrow {\log }_{3}2+{\log }_3(2^x-\frac{7}{2})=2\times {\log }_3(2^x-5)$
$\Rightarrow {\log }_3(2\times (2^x-\frac{7}{2}\left)\right)={\log }_3(2^x-5{)}^2$
$\Rightarrow 2^{x+1}-7=(2^x-5{)}^2$
i.e. $2^{x+1}-7=2^{2x}-5\times 2^{x+1}+25$
i.e. $2^{2x}-6\times 2^{x+1}+32=0$
i.e. $2^{2x}-12\times 2^x+32=0$
i.e. $(2^x-4)\times (2^x-8)=0$
i.e. $2^x=4,8orx=2,3$
However, when $x$ is $2,2^x-5$ becomes $-1$ which is not allowed inside logarithm.
Hence $x=3.$