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##

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