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

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Answer 1 · 2015-12-01T11:12:46

$- 6 - 4 t$ $-6-4t$

${ln (\frac{1}{e})}^{6} - 4 t = {ln e}^{- 6} - 4 t$ $ln(1/e)^6-4t=lne^-6-4t$
$= - 6 ln e - 4 t$ $=-6lne-4t$
$= - 6 - 4 t$ $=-6-4t$

Answer 2 · 2015-12-01T11:13:51

$- 2 (3 + 2 t)$ $-2(3+2t)$

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

${ln (\frac{1}{e})}^{6} = ln (e^{- 6})$ $ln(1/e)^6 = ln(e^-6)$

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

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

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

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