SOLUTION: solve the logarithmic equation algebraically. e^x + e^-x = 4 please explain how you do this in detail. Thank you!

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Question 1028474: solve the logarithmic equation algebraically.

e^x + e^-x = 4
please explain how you do this in detail. Thank you!
Found 2 solutions by Alan3354, Edwin McCravy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
solve the logarithmic equation algebraically.

e^x + e^-x = 4
----------
Multiply thru by e^2
e^(2x) + 1 = 4e^x
e^(2x) - 4e^x + 1 = 0
Now it's quadratic in e^x.
-------------
Or,
e^x + e^-x = 4
(e^x + e^-x)/2 = 2
sinh(x) = 2
x =~ 1.443635

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!

e%5Ex+%2B+e%5E%28-x%29%22%22=%22%224

A negative exponent means one over the base with the
positive exponent:

e%5Ex+%2B+1%2Fe%5Ex%22%22=%22%224

Multiply through by e%5E%28x%29

e%5E%282x%29+%2B+1%22%22=%22%224e%5Ex

e%5E%282x%29+-+4e%5Ex+%2B+1%22%22=%22%220

%28e%5Ex%29%5E2+-+4e%5Ex+%2B+1%22%22=%22%220

Let u=e%5Ex

u%5E2-4u%2B1%22%22=%22%220

 u+=+%28-b+%2B-+sqrt%28+b%5E2-4ac+%29%29%2F%282a%29+

 u+=+%28-%28-4%29+%2B-+sqrt%28+%28-4%29%5E2-4%281%29%281%29+%29%29%2F%282%281%29%29+

 u+=+%284+%2B-+sqrt%2816-4+%29%29%2F2+

 u+=+%284+%2B-+sqrt%2812%29%29%2F2+

 u+=+%284+%2B-+sqrt%284%2A3%29%29%2F2+

 u+=+%284+%2B-+2sqrt%283%29%29%2F2+

 u+=+%282%282+%2B-+sqrt%283%29%29%29%2F2+

 u+=+%28cross%282%29%282+%2B-+sqrt%283%29%29%29%2Fcross%282%29+

 u+=+2+%2B-+sqrt%283%29+

Since u=e%5Ex

 e%5Ex+=+2+%2B-+sqrt%283%29+

  x+=+ln%282+%2B-+sqrt%283%29%29+

x=%22%22+%2B-+1.316957897


The second way the other tutor did it, he mistakenly 
wrote sinh for cosh (hyperbolic functions) and got the 
wrong solution.  It should have been

cosh(x) = 2

which does gives the answer above:

x=%22%22+%2B-+1.316957897

Edwin