How do you solve $\frac{400}{1 + e^{- x}} = 350$ $400/(1+e^-x)=350$ ?

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Answer 1 · 2017-01-04T06:52:14

ln 7

Cross multiplying and rearranging,

$r^{- x} = \frac{1}{7}$ $r^(-x)=1/7$ . The reciprocal

$e^{x} = 7$ $e^x=7$ .

Inversely,

$x = ln 7$ $x = ln 7$ .

How do you solve 4001+e−x=350400/(1+e^-x)=350?