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When I evaluate ln(1/e) it comes out as -1, I'm confused how as the steps don't make sense to me. Do you use logb(m/n)=logbM-logbN? I thought ln and e would cancel and you'd be left with -1.

 Jan 11, 2019

Best Answer 

 #1
avatar+5225 
+2

\(\dfrac 1 e = e^{-1}\\ \ln\left(e^{-1}\right) = -1\\ \ln\left(\dfrac 1 e\right) = -1\)

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 Jan 11, 2019
 #1
avatar+5225 
+2
Best Answer

\(\dfrac 1 e = e^{-1}\\ \ln\left(e^{-1}\right) = -1\\ \ln\left(\dfrac 1 e\right) = -1\)

Rom Jan 11, 2019
 #2
avatar+101778 
+1

Remember that    1 / e   =  e^(-1)

 

So.....

 

ln (1 / e)   =   ln e^(-1)    

 

And by  a log property.....   ln a^b   =  b * ln a

 

So

 

ln e^(-1)  =   -1 * ln  e

 

But ln e   =  1   .....so

 

(-1) * (1)  =   -1

 

 

cool cool cool

 Jan 11, 2019

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