##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.
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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##
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##color(blue)(Footnote)##
In conclusion the general rule applies: ## a^(log_a(x))=x##