



The
twin of exponential equations are logarithms.



If we say
... 


Y = e^{X}



That also
means ... 


log _{e}Y
= X



(also
known as ln Y = X)






Remember
the way you read a log is: 








Which is
the same thing we said in the beginning ... 


Y = e^{X}






(NOTE:
From here on we'll be using ln instead of log_{e})






So the
point is: 


Y
= e^{X }and ln Y = X



Say the
exact same thing. 





So let's
talk about the derivatives and integrals of things like 





ln
X






When we
went looking for the derivative of e^{X}, we found that it
was e^{X}, 


give or
take a dX. 


Wouldn't
it be nice if ln X worked the same way 


SORRY, No
such luck. 


But
finding it isn't too bad. 





Start
with 


Y
= ln X






which is
the same thing as saying: 


X =
e^{Y}



(Don't
worry, we'll be back to ln X before we're done)






OK, the
derivative of e^{Y} (with respect to the variable Y) is e^{Y} 


so we can
say: 








We can
flip these over and say: 








Now we do
some substitution. 


We
started with the equations: 





Y
= ln X also known as X =
e^{Y}






Now we
substitute both of these into our last equation: 











In
English, what this says is that the derivative of lnX, 


is equal to
1/X (also known as X^{1}) 


This also
means that the integral of X^{1} is lnX, 


remembering,
of course, to include dX's and + C's where needed. 











Sometimes
you will see written as 


Don't let
that throw you, they both mean the same thing. 





Examples: 








Did you
notice the 2X in the numerator of the last one? 


The
derivative of X^{2} is 2XdX, so everything we needed
to have, 


to
"Put Back Into" the 1/X^{2} was there. 





So this
all works out great for things like e^{X} and ln X, 


but what
if you have to deal with 2^{X} or log_{8}X ? 


Huh??? 


Well????? 





copyright 2005 Bruce Kirkpatrick 
