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

Dec 1, 2015

$- 6 - 4 t$

Explanation:

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

Dec 1, 2015

$- 2 \left(3 + 2 t\right)$

Explanation:

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

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

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

And by definition, $\ln \left(e\right) = 1$

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