Question 185080
{{{2e^(x-2) = e^x + 7 }}} Start with the given equation.



{{{2e^(x)/e^2 = e^x + 7}}} Rewrite {{{e^(x-2)}}} as {{{e^(x)/e^2}}} using the identity {{{x^(y-z)=x^y/x^z}}} where {{{x<>0}}}



{{{2e^(x) = e^x*e^2 + 7*e^2}}} Multiply EVERY term by the LCD {{{e^2}}} to get rid of the fraction.



{{{2e^(x) - e^x*e^2 = 7*e^2}}} Subtract {{{e^x*e^2}}} from both sides.



{{{e^(x)(2 - e^2) = 7*e^2}}} Factor out the GCF {{{e^x}}}



{{{e^(x) = 7*e^2/(2 - e^2)}}} Divide both sides by {{{2 - e^2}}}.



{{{e^(x) = -9.597}}} Evaluate the right side with a calculator. 



{{{x = ln(-9.597)}}} Take the natural log of both sides (to eliminate the base "e")



Since you CANNOT take the natural log of a negative number (well at least for now...), this means that there is no solution. 




So there are no solutions. I would double check the original problem.



As visual confirmation, graph the two expressions {{{2e^(x-2)}}} and {{{e^x + 7}}} and you'll find that they do NOT cross. 




Note: taking the log of a negative number will give you a complex number, but that's extra information...