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# How do you simplify ##e^-lnx##?

##e^(-ln(x))" " =" " 1/x##

##color(brown)("Total rewrite as changed my mind about pressentation.")##

##color(blue)("Preamble:")##

Consider the generic case of ##" "log_10(a)=b##

Another way of writing this is ##10^b=a##

Suppose ##a=10 ->log_10(10)=b##

##=>10^b=10 => b=1##

So ##color(red)(log_a(a)=1 larr" important example")##

We are going to use this principle. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Write ##" "e^(-ln(x))" "## as ##" "1/(e^(ln(x))##

Let ##y=e^(ln(x)) =>" "1/y=1/(e^(ln(x))## ..................Equation(1)

....................................................................................... Consider just the denominators and take logs of both sides

##y=e^(ln(x))" " ->" "ln(y)=ln(e^(ln(x)))##

But for generic case ##ln(s^t) -> tln(s)##

##color(green)(=>ln(y)=ln(x)ln(e))##

But ##log_e(e)" "->" "ln(e)=1 color(red)(larr" from important example")##

##color(green)(=>ln(y)=ln(x)xx1)##

Thus ##y=x## .....................................................................................

So Equation(1) becomes

##1/y" "=" "1/(e^(ln(x)))" "=" "1/x##

Thus ##e^(-ln(x)) = 1/x##

~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ##color(blue)("Footnote")##

In conclusion the general rule applies: ##" "a^(log_a(x))=x##