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)   (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)   (Show Source): You can put this solution on YOUR website!



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



Multiply through by 







Let 



 

 

 

 

 

 

 

 

 

Since 

 

  




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:



Edwin

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