How do you solve #-8e^(-p)=-49#?

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Answer 1 · 2018-06-20T16:05:18

#p = log(8/49)#

Divide both sides by #-8#:

#e^{-p} = 49/8#

consider logarithm of both sides:

#log(e^{-p}) = log(49/8)#

since #log(e^s) = s#, we have

#-p = log(49/8)#

and thus

#p = -log(49/8) = log(8/49)#