Question 934515
{{{e^x+e^(2x)=12}}}

{{{e^x+e^x*e^x=12}}}

{{{e^x*(e^x+1) =3*4}}}

if product same then corresponding factors are same too; so, we have

{{{e^x= 3}}}

and

{{{e^x+1 = 4}}}=>{{{e^x = 4-1}}}=>{{{e^x= 3}}}

{{{log(e^x)= log(3)}}}

{{{x*log(e)= log(3)}}}......{{{log(e)=1}}}

{{{x = log(3)}}}