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)#