Question

if log 2,log(2^x-1) and log (2^x-3) are in AP . find x

Answer 1

given

$log(2) \: log( {2}^{x} - 1) \: log( {2}^{x} - 3 ) \: \: are \: in \: ap$
so

$2 log( {2}^{x} - 1) = log(2) + log( {2}^{x} - 1 ) \\ \\2 log( {2}^{x} - 1) = log(2( {2}^{x} - 1) \\ log( ({2x - 1})^{2} ) = log(2( {2}^{x} - 1) \\ \\ taking \: antilog \: both \: sides \\ { ({2}^{x} - 1)}^{2} = 2( {2}^{x} - 1) \\ {2}^{2x} + 1 - {2}^{x } \times 2 = 2 \times {2}^{x} - 2 \\ now \: take \: {2}^{x} = t \\ and \: solve \: quadratic \: eq$
hope it will help you!