# Thread: May I have a question?

1. ## 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.

2. ## Re: May I have a question?

It is giving all complex solutions to the problem. Given any complex number, $x= r(cos(\theta)+ i sin(\theta)= re^{i\theta}$, $ln(x)= ln(r)+ i\theta+ 2i n\pi$, the " $+ 2i n\pi$ giving equivalent values.

3. ## Re: May I have a question?

Hi Master

Then, may I ask if I was wrong if I just put 700*(ln5/ln2) in terms of the question's answer?

4. ## Re: May I have a question?

Originally Posted by yanirose
Hi Master

Then, may I ask if I was wrong if I just put 700*(ln5/ln2) in terms of the question's answer?
It depends whether the question explicitly asked for a real solution or implicitly restricted the answer to the domain of the reals.

5. ## Re: May I have a question?

Originally Posted by yanirose
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.
Is the equation \displaystyle \begin{align*} e^{-x\ln{ \left( \frac{2}{700} \right) }} \end{align*} or \displaystyle \begin{align*} e^{\frac{-x\ln{(2)}}{700}} \end{align*}?

6. ## Re: May I have a question?

Dear Masters

The questions is the latter.
This solution was to ask about the decay proportion, so the x is years

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

8. ## Re: May I have a question?

Thanks. so I got the answer for x = 700*(ln5/ln2). And I left the answer as it is because I did not have any approximation value for the log.
I was worrying if I am wrong because the mathematica has a complex value