How do you simplify $e^{- ln x}$ $e^-lnx$ ?

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Answer 1 · 2016-02-06T15:01:51

$e^{- ln (x)} = \frac{1}{x}$ $e^(-ln(x))" " =" " 1/x$

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

$Preamble:$ $color(blue)("Preamble:")$

Consider the generic case of ${log}_{10} (a) = b$ $" "log_10(a)=b$

Another way of writing this is $10^{b} = a$ $10^b=a$

Suppose $a = 10 \to {log}_{10} (10) = b$ $a=10 ->log_10(10)=b$

$\Rightarrow 10^{b} = 10 \Rightarrow b = 1$ $=>10^b=10 => b=1$

So ${log}_{a} (a) = 1 \leftarrow important example$ $color(red)(log_a(a)=1 larr" important example")$

We are going to use this principle.
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Write $e^{- ln (x)}$ $" "e^(-ln(x))" "$ as $\frac{1}{e^{ln (x)}}$ $" "1/(e^(ln(x))$

Let $y = e^{ln (x)} \Rightarrow \frac{1}{y} = \frac{1}{e^{ln (x)}}$ $y=e^(ln(x)) =>" "1/y=1/(e^(ln(x))$ ..................Equation(1)

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Consider just the denominators and take logs of both sides

$y = e^{ln (x)} \to ln (y) = ln (e^{ln (x)})$ $y=e^(ln(x))" " ->" "ln(y)=ln(e^(ln(x)))$

But for generic case $ln (s^{t}) \to t ln (s)$ $ln(s^t) -> tln(s)$

$\Rightarrow ln (y) = ln (x) ln (e)$ $color(green)(=>ln(y)=ln(x)ln(e))$

But ${log}_{e} (e) \to ln (e) = 1 \leftarrow from important example$ $log_e(e)" "->" "ln(e)=1 color(red)(larr" from important example")$

$\Rightarrow ln (y) = ln (x) \times 1$ $color(green)(=>ln(y)=ln(x)xx1)$

Thus $y = x$ $y=x$
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So Equation(1) becomes

$\frac{1}{y} = \frac{1}{e^{ln (x)}} = \frac{1}{x}$ $1/y" "=" "1/(e^(ln(x)))" "=" "1/x$

Thus $e^{- ln (x)} = \frac{1}{x}$ $e^(-ln(x)) = 1/x$

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$Footnote$ $color(blue)("Footnote")$

In conclusion the general rule applies: $a^{{log}_{a} (x)} = x$ $" "a^(log_a(x))=x$

How do you simplify e−lnxe^-lnx?