Question #2641d

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Answer 1 · 2018-02-06T14:36:12

$1/x$

Remember that $a^b=e^(b*lna)$ .

Here, $a=x$ , and $b=-1$ .

The above equation becomes $e^(-1*lnx)$ .

Using the above property of powers, this simplifies to:

$x^-1$

$=1/x$

Answer 2 · 2018-02-06T14:38:27

The answer is $=1/x$

Let $y=e^(-lnx)$

Taking the logarithms on both sides,

$lny=ln(e^(-lnx))$

$=-lnx$

$=ln1-lnx$

$=ln(1/x)$

As,

$lny=ln(1/x)$

$y=1/x$