How do you solve $- 8 e^{- p} = - 49$ $-8e^(-p)=-49$ ?

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Answer 1 · 2018-06-20T16:05:18

$p = log (\frac{8}{49})$ $p = log(8/49)$

Divide both sides by $- 8$ $-8$ :

$e^{- p} = \frac{49}{8}$ $e^{-p} = 49/8$

consider logarithm of both sides:

$log (e^{- p}) = log (\frac{49}{8})$ $log(e^{-p}) = log(49/8)$

since $log (e^{s}) = s$ $log(e^s) = s$ , we have

$- p = log (\frac{49}{8})$ $-p = log(49/8)$

and thus

$p = - log (\frac{49}{8}) = log (\frac{8}{49})$ $p = -log(49/8) = log(8/49)$

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