Question 1189604
Find an equation of the tangent line to the graph of the function at the given point:

{{{1+ ln (7xy) = e^(7x-y)}}}, ({{{1/7}}},{{{1}}})


{{{1+ ln (7xy) - e^(7x-y)=0}}}


{{{log(x) + log(y) + log(7)- e^(7x - y)=0}}}


{{{(d/dx)(log(x) + log(y) + log(7))=1/x}}}


{{{(d/dx)(e^(7 x - y))= 7 e^(7 x - y)}}}


{{{y}}}'{{{(x)=1/x-7e^(7x - y)}}}



evaluate at ({{{1/7}}},{{{1}}})


{{{ 1/(1/7)-7 e^((7(1/7)) - 1)=7-7e^(1-1) =7-7*e^0=7-7*1=0}}}


slope of tangent is {{{m=0}}}, so the line will be of the form {{{y=a}}}

 given point ({{{1/7}}},{{{1}}}), and tangent line is

{{{y=1}}}


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