changing the log expression to equal expression with exponent

**blondedude092** · May 1st 2012, 04:03 AM

does ln 1/e^6 = -6

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

**biffboy** · May 1st 2012, 04:11 AM

Correct

**Prove It** · May 1st 2012, 05:12 AM

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

**HallsofIvy** · May 1st 2012, 07:55 AM

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.

**biffboy** · May 1st 2012, 08:03 AM

As an alternative we could say ln1/e^6 = ln1-lne^6 =0-6 = -6

**earboth** · May 1st 2012, 09:54 AM

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

Thread: changing the log expression to equal expression with exponent