Question 484241
{{{e^(x)+e^(-x)=4}}} Start with the given equation.



{{{e^(x)+1/(e^(x))=4}}} Rewrite {{{e^(-x)}}} as {{{1/(e^(x))}}}



{{{e^(2x)+1=4e^x}}} Multiply EVERY term by the LCD {{{e^x}}} to clear out the fractions.



{{{e^(2x)+1-4e^x=0}}} Get everything to one side.



{{{e^(2x)-4e^x+1=0}}} Rearrange the terms.



{{{(e^x)^2-4e^x+1=0}}} Rewrite the first term



{{{z^2-4z+1=0}}} Replace EVERY copy of {{{e^x}}} with 'z' (so {{{z=e^x}}})



Notice that the quadratic {{{z^2-4z+1}}} is in the form of {{{Az^2+Bz+C}}} where {{{A=1}}}, {{{B=-4}}}, and {{{C=1}}}



Let's use the quadratic formula to solve for "z":



{{{z = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{z = (-(-4) +- sqrt( (-4)^2-4(1)(1) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-4}}}, and {{{C=1}}}



{{{z = (4 +- sqrt( (-4)^2-4(1)(1) ))/(2(1))}}} Negate {{{-4}}} to get {{{4}}}. 



{{{z = (4 +- sqrt( 16-4(1)(1) ))/(2(1))}}} Square {{{-4}}} to get {{{16}}}. 



{{{z = (4 +- sqrt( 16-4 ))/(2(1))}}} Multiply {{{4(1)(1)}}} to get {{{4}}}



{{{z = (4 +- sqrt( 12 ))/(2(1))}}} Subtract {{{4}}} from {{{16}}} to get {{{12}}}



{{{z = (4 +- sqrt( 12 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{z = (4 +- 2*sqrt(3))/(2)}}} Simplify the square root  



{{{z = (4)/(2) +- (2*sqrt(3))/(2)}}} Break up the fraction.  



{{{z = 2 +- sqrt(3)}}} Reduce.  



{{{z = 2+sqrt(3)}}} or {{{z = 2-sqrt(3)}}} Break up the expression.  



So the solutions are {{{z = 2+sqrt(3)}}} or {{{z = 2-sqrt(3)}}} 



But remember, we said that {{{z=e^x}}}



So {{{e^x = 2+sqrt(3)}}} or {{{e^x = 2-sqrt(3)}}}



Now convert to logarithmic form to get 


{{{x = ln(2+sqrt(3))}}} or {{{x = ln(2-sqrt(3))}}}



So the solutions are 


{{{x = ln(2+sqrt(3))}}} or {{{x = ln(2-sqrt(3))}}}