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5 answers

=ln x +-x= ln(x-x)

2007-12-09 06:34:14 · answer #1 · answered by star baller 360 5 · 0 3

ln x - ln e^x (because ln (a/b) = ln a - ln b)

ln x - x ( because ln a^b = b ln a ; and ln e =1 )

i think that is what you are asking for

2007-12-09 14:28:44 · answer #2 · answered by michael c 3 · 1 0

= ln x-xln e and as ln e =1
=ln x -x

2007-12-09 14:32:19 · answer #3 · answered by santmann2002 7 · 1 0

ln(x) - ln(e^x)

ln(x) - x*ln(e)

ln(x) - x*1

ln(x) - x

2007-12-09 14:25:37 · answer #4 · answered by ben e 7 · 2 0

ln (x/e^x) = ln(x * e^(-x)) = ln(x) + ln(e^(-x)) = ln(x) + (-x) = ln(x) - x

2007-12-09 14:25:06 · answer #5 · answered by antone_fo 4 · 2 0

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