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How do you simplify #e^-lnx#?

1 Answer
Feb 6, 2016

Answer:

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

Explanation:

#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#