Question

How do you simplify `e^-lnx`?

Answer 1

`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`