changing the log expression to equal expression with exponent

does ln 1/e^6 = -6

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

Re: changing the log expression to equal expression with exponent

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 $\displaystyle \displaystyle \begin{align*} \ln{\left(\frac{1}{e^6}\right)} \end{align*}$, yes you are correct :)

Re: changing the log expression to equal expression with exponent

In general, ln(a)= b is the same as (converts into) $\displaystyle e^b= a$ so, in particular, $\displaystyle ln(1/e^{6})= -6$, $\displaystyle 1/e^6= e^{-6}$ because $\displaystyle a= 1/e^6$ and b= -6.

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

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