# How do you simplify e^-lnx?

Feb 6, 2016

${e}^{- \ln \left(x\right)} \text{ " =" } \frac{1}{x}$

#### Explanation:

$\textcolor{b r o w n}{\text{Total rewrite as changed my mind about pressentation.}}$

$\textcolor{b l u e}{\text{Preamble:}}$

Consider the generic case of $\text{ } {\log}_{10} \left(a\right) = b$

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

Suppose $a = 10 \to {\log}_{10} \left(10\right) = b$

$\implies {10}^{b} = 10 \implies b = 1$

So $\textcolor{red}{{\log}_{a} \left(a\right) = 1 \leftarrow \text{ important example}}$

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

Let 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

$y = {e}^{\ln \left(x\right)} \text{ " ->" } \ln \left(y\right) = \ln \left({e}^{\ln \left(x\right)}\right)$

But for generic case $\ln \left({s}^{t}\right) \to t \ln \left(s\right)$

$\textcolor{g r e e n}{\implies \ln \left(y\right) = \ln \left(x\right) \ln \left(e\right)}$

But log_e(e)" "->" "ln(e)=1 color(red)(larr" from important example")

$\textcolor{g r e e n}{\implies \ln \left(y\right) = \ln \left(x\right) \times 1}$

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

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

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

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$\textcolor{b l u e}{\text{Footnote}}$

In conclusion the general rule applies: $\text{ } {a}^{{\log}_{a} \left(x\right)} = x$