How do you solve ${ln e}^{- 2 x} = 8$ $ln e^(-2x) = 8$ ?

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Answer 1 · 2016-02-19T11:34:12

I found: $x = - 4$ $x=-4$

We can first take the $- 2 x$ $-2x$ and place it in front of the log:

$- 2 x ln e = 8$ $-2xlne=8$

then remember that: $ln e = 1$ $lne=1$
and write:

$- 2 x = 8$ $-2x=8$
so that $x = - \frac{8}{2} = - 4$ $x=-8/2=-4$

How do you solve lne−2x=8ln e^(-2x) = 8?