How do you solve #7e^{x} = 9- e^{- x}#?

1 Answer
Salvatore I.
Nov 13, 2016

Two solutions
#x_1=ln((9-root2(53))/14)#
#x_2=ln((9+root2(53))/14)#

Explanation:

Let's rewrite the equation
#7e^x-9+1/e^x=0#
We can multiply both sides for #e^x# so that we get

#7e^(2x)-9e^x+1=0#
this can be solved as an ordinary second degree eqaution

#e^x=(9+-root2(81-28))/14#

and consequently

#x=ln((9+-root2(53))/14)#

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