If $${\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right),{\log _{10}}\left( {{2^x} + 3} \right)$$       are three consecutive terms of an A.P., then which one of the following is correct ?

Solution :
$$\eqalign{ & {\text{Let, }}{\log _{10}}2,{\log _{10}}\left( {{2^x} - 1} \right){\text{and }}{\log _{10}}\left( {{2^x} + 3} \right){\text{are in A}}{\text{.P}}{\text{.}} \cr & \therefore 2{\log _{10}}\left( {{2^x} - 1} \right) = {\log _{10}}2 + {\log _{10}}\left( {{2^x} + 3} \right) \cr & \Rightarrow {\log _{10}}{\left( {{2^x} - 1} \right)^2} = {\log _{10}}2\left( {{2^x} + 3} \right) \cr & \Rightarrow {2^{2x}} + 1 - {2^{x + 1}} = {2.2^x} + 6 \cr & \Rightarrow {a^2} + 1 - 2a = 2a + 6\,\,\,\,{\text{where }}a = {2^x}. \cr & \Rightarrow {a^2} - 4a - 5 = 0 \cr & \Rightarrow a = 5\,\,{\text{or }}a = - 1 \cr & {2^x} = 5 \cr & \Rightarrow \log 2 = \log 5 \cr & \Rightarrow x = \frac{{\log 5}}{{\log 2}} \cr & \Rightarrow x = {\log _2}5 \cr} $$