How do you solve #3000/(2+e^(2x))=2#?

1 Answer
sente
Dec 13, 2015

#x= ln(1498)/2#

Explanation:

We will use the following properties of logarithms:

  • #ln(a^x) = xln(a)#

  • #log_a(a) = 1#

#3000/(2+e^(2x)) = 2#

#=> 3000 = 2(2+e^(2x))=4 + 2e^(2x)#

#=> 2996 = 2e^(2x)#

#=> 1498 = e^(2x)#

#=> ln(1498) = ln(e^(2x)) = 2xln(e) = 2x(1) = 2x#

#=> x = ln(1498)/2#

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