$\text{ If }{\log }_5^2\text{,}{\log }_5^{\left(2^x-3\right)},{\log }_5^{\left(\frac{17}{2}+2^{x-1}\right)}\text{ are in A.P then }\mathrm{x}\text{ is }$

$2{\log }_5^{\left(2^x-3\right)}=$${\log }_5^2 +{\log }_5^{\left(\frac{17}{2}+2^{x-1}\right)}$

$\begin{array}{l}\Rightarrow {\left(2^x-3\right)}^2=\left(17+2^{\mathrm{x}}\right)\\ \text{ Let }{2}^{\mathrm{x}}=\mathrm{y}\\ \Rightarrow {\mathrm{y}}^2-7\mathrm{y}-8=0\to \mathrm{y}=8_{ }or y=-1, which is not possible \sin ce y>0\\ \text{ If }\mathrm{y}=8\to 2^{\mathrm{x}}=8\\ \mathrm{x}=3.\end{array}$