How do you simplify $ln(1/e)^6-4t$ ?

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Answer 1 · 2015-12-01T11:12:46

$-6-4t$

$ln(1/e)^6-4t=lne^-6-4t$
$=-6lne-4t$
$=-6-4t$

Answer 2 · 2015-12-01T11:13:51

$-2(3+2t)$

First of all, we have that $log(x^a)=alog(x)$ . And since $1/e=e^{-1}$ , we have that

$ln(1/e)^6 = ln(e^-6)$

So, $ln(e^{-6}) = -6ln(e)$

And by definition, $ln(e)=1$

So, the expression becomes $-6-4t=-2(3+2t)$

How do you simplify ln(1/e)^6-4tln(1/e)^6-4t?