Question 998703
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Solve e^2x-10e^x+24=0
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Factor


{{{e^2x-10e^x+24}}} = {{{(e^x-4)*(e^x-6)}}}.


Now the roots are 


{{{e^x}}} = {{{4}}}   ----->  x = {{{ln(4)}}},     and


{{{e^x}}} = {{{6}}}   ----->  x = {{{ln(6)}}}.