It is giving allcomplexsolutions to the problem. Given any complex number, , , the " giving equivalent values.

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- May 3rd 2014, 05:17 AM #1

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## May I have a question?

Hi Masters,

I calculated 0.2 = e^(-x(ln2/700)) to find x.

But, I could not use a calculator also did not have any approximation for the value of ln.

So, I got x = 700(ln5 / ln2).

But the wolfram says the answer is x = 700((ln5+ 2i pi n) / ln2).

Can I ask where 2i pi n came from?

Was I wrong?

Thanks a lot.

- May 3rd 2014, 05:58 AM #2

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- May 3rd 2014, 07:14 AM #3

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- May 3rd 2014, 05:55 PM #4

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- May 3rd 2014, 07:09 PM #5

- May 4th 2014, 05:20 AM #6

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- May 4th 2014, 05:47 AM #7
## Re: May I have a question?

Well for starters, $\displaystyle \begin{align*} \frac{-x\ln{(2)}}{700} = -\frac{x}{700}\ln{(2)} = \ln{ \left( 2^{-\frac{x}{700}} \right) } \end{align*}$, so that means $\displaystyle \begin{align*} e^{-\frac{x\ln{(2)}}{700}} = e^{\ln{ \left( 2^{-\frac{x}{700}} \right) }} = 2^{-\frac{x}{700}} \end{align*}$

I think $\displaystyle \begin{align*} 0.2 = 2^{-\frac{x}{700}} \end{align*}$ is much easier to solve

- May 4th 2014, 08:29 AM #8

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