Question

Log7(2^x-1) + log7(2^x-7) =1

Answer 1

its will be as follows,,,

$log_7 ( {2}^{x} - 1) + log_{7}( {2}^{x} - 7 ) = 1$

$log_{7}(( {2}^{x} - 1) ( {2}^{x} - 7)) = 1$

By defination,

${2}^{2x} - 8. {2}^{x} + 7 = {7}^{1}$

${2}^{2x} - {2}^{3} . {2}^{x} = 0$

${2}^{2x} - {2}^{3 + x} = 0$

${2}^{2x} = {2}^{3 + x}$

As the bases are same the indices will be same ,

$2x = 3 + x$

$x = 3$

Hope it helped you

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Answer 2

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