# changing the log expression to equal expression with exponent

• May 1st 2012, 05:03 AM
blondedude092
changing the log expression to equal expression with exponent
does ln 1/e^6 = -6

convert into 1/e^6 = e ^-6 ???
• May 1st 2012, 05:11 AM
biffboy
Re: changing the log expression to equal expression with exponent
Correct
• May 1st 2012, 06:12 AM
Prove It
Re: changing the log expression to equal expression with exponent
Quote:

Originally Posted by blondedude092
does ln 1/e^6 = -6

convert into 1/e^6 = e ^-6 ???

Assuming you were trying to write \displaystyle \begin{align*} \ln{\left(\frac{1}{e^6}\right)} \end{align*}, yes you are correct :)
• May 1st 2012, 08:55 AM
HallsofIvy
Re: changing the log expression to equal expression with exponent
In general, ln(a)= b is the same as (converts into) $e^b= a$ so, in particular, $ln(1/e^{6})= -6$, $1/e^6= e^{-6}$ because $a= 1/e^6$ and b= -6.
• May 1st 2012, 09:03 AM
biffboy
Re: changing the log expression to equal expression with exponent
As an alternative we could say ln1/e^6 = ln1-lne^6 =0-6 = -6
• May 1st 2012, 10:54 AM
earboth
Re: changing the log expression to equal expression with exponent
Quote:

Originally Posted by biffboy
As an alternative we could say ln1/e^6 = ln1-lne^6 =0-6 = -6

I don't want to pick at you but to be exact

ln1/e^6 = 0