SOLUTION: I'm having trouble with this problem. Thanks for your help. Solve. Express x as a logarithm if necessary. e^(x)+e^(-x)=4

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I'm having trouble with this problem. Thanks for your help. Solve. Express x as a logarithm if necessary. e^(x)+e^(-x)=4      Log On


   

Question 484241: I'm having trouble with this problem. Thanks for your help.
Solve. Express x as a logarithm if necessary.
e^(x)+e^(-x)=4
Found 2 solutions by jim_thompson5910, chessace:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
e%5E%28x%29%2Be%5E%28-x%29=4 Start with the given equation.


e%5E%28x%29%2B1%2F%28e%5E%28x%29%29=4 Rewrite e%5E%28-x%29 as 1%2F%28e%5E%28x%29%29


e%5E%282x%29%2B1=4e%5Ex Multiply EVERY term by the LCD e%5Ex to clear out the fractions.


e%5E%282x%29%2B1-4e%5Ex=0 Get everything to one side.


e%5E%282x%29-4e%5Ex%2B1=0 Rearrange the terms.


%28e%5Ex%29%5E2-4e%5Ex%2B1=0 Rewrite the first term


z%5E2-4z%2B1=0 Replace EVERY copy of e%5Ex with 'z' (so z=e%5Ex)


Notice that the quadratic z%5E2-4z%2B1 is in the form of Az%5E2%2BBz%2BC where A=1, B=-4, and C=1


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


z+=+%28-B+%2B-+sqrt%28+B%5E2-4AC+%29%29%2F%282A%29 Start with the quadratic formula


z+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%281%29%281%29+%29%29%2F%282%281%29%29 Plug in A=1, B=-4, and C=1


z+=+%284+%2B-+sqrt%28+%28-4%29%5E2-4%281%29%281%29+%29%29%2F%282%281%29%29 Negate -4 to get 4.


z+=+%284+%2B-+sqrt%28+16-4%281%29%281%29+%29%29%2F%282%281%29%29 Square -4 to get 16.


z+=+%284+%2B-+sqrt%28+16-4+%29%29%2F%282%281%29%29 Multiply 4%281%29%281%29 to get 4


z+=+%284+%2B-+sqrt%28+12+%29%29%2F%282%281%29%29 Subtract 4 from 16 to get 12


z+=+%284+%2B-+sqrt%28+12+%29%29%2F%282%29 Multiply 2 and 1 to get 2.


z+=+%284+%2B-+2%2Asqrt%283%29%29%2F%282%29 Simplify the square root


z+=+%284%29%2F%282%29+%2B-+%282%2Asqrt%283%29%29%2F%282%29 Break up the fraction.


z+=+2+%2B-+sqrt%283%29 Reduce.


z+=+2%2Bsqrt%283%29 or z+=+2-sqrt%283%29 Break up the expression.


So the solutions are z+=+2%2Bsqrt%283%29 or z+=+2-sqrt%283%29


But remember, we said that z=e%5Ex


So e%5Ex+=+2%2Bsqrt%283%29 or e%5Ex+=+2-sqrt%283%29


Now convert to logarithmic form to get

x+=+ln%282%2Bsqrt%283%29%29 or x+=+ln%282-sqrt%283%29%29


So the solutions are

x+=+ln%282%2Bsqrt%283%29%29 or x+=+ln%282-sqrt%283%29%29



Answer by chessace(471) About Me  (Show Source):
You can put this solution on YOUR website!
Let z=e%5Ex
Note that y%5E%28-a%29 = 1%2Fy%5Ea
So we have z + 1/z = 4
*z: z%5E2 + 1 = 4z
z%5E2 - 4z +1 =0
z = %28-%28-4%29%2B-sqrt%28%28-4%29%5E2-4%281%29%281%29%29%29%2F%282%2A1%29
z = %284%2B-sqrt%2816-4%29%29%2F2
sqrt%2812%29 = sqrt%284%2A3%29 = 2 * sqrt(3)
z = 2%2B-sqrt%283%29%29
Ln(z) = x (natural log base e)
So x = Ln%282+%2B-+sqrt%283%29%29