SOLUTION: Hi, Can you please help, . Solve {{{ e^x + e^(-x) = 5 }}}, for "x" . I would like some answers quickly, so I can put it down for notes for a quiz tommorrow. . Thanks ahead of

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hi, Can you please help, . Solve {{{ e^x + e^(-x) = 5 }}}, for "x" . I would like some answers quickly, so I can put it down for notes for a quiz tommorrow. . Thanks ahead of      Log On


   

Question 168634: Hi, Can you please help,
.
Solve +e%5Ex+%2B+e%5E%28-x%29+=+5+, for "x"
.
I would like some answers quickly, so I can put it down for notes for a quiz tommorrow.
.
Thanks ahead of time, Levi
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
e^x + e^(-x) = 5
e^x + 1/e^x = 5
Multiply thru by e^x to get:
e^2x + 1 = 5 e^x
Rearrange to form a quadratic equation:
(e^x)2 - 5e^x + 1 = 0
---------------------------
Change the variable: Let w = e^x
Rewrite:
w^2 - 5w + 1 = 0
Use the quadratic formula:
w = [5 +- sqrt(25-4*1)]/2
w = [5 +- sqrt(21)]/2
--------------------------
Change back to e^x
e^x = [5 + sqrt(21)]/2 or e^x = [5 - sqrt(21)]/2
Take the natural log of both sides to get:
x = ln[[5 + sqrt(21)]/2] or x = ln[[5 - sqrt(21)]/2]
x = ln[4.79129] or x = ln[208712]
x = 1.5668 or x = -1.5668
=============================
Cheers,
Stan H.