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March 30, 2015

March 30, 2015

Posted by **Alli** on Friday, January 30, 2009 at 10:40pm.

1) e^(2(ln)(x))

2) e^(-lnx)

I know the rules I just dont understand how they apply..

thanks!

- Trig -
**Reiny**, Friday, January 30, 2009 at 10:53pmthe "rule" here is that

e^ln(k) = k

so e^(2lnx)

= e^ln(x^2)

= x^2

and

2) e^(-lnx)

= e^ln(x^-1)

= x^-1

or 1/x

- Trig -
**drwls**, Saturday, January 31, 2009 at 3:41pmThis is not the trigonometry I studied in high school

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