# Natural log and smallest number

• Nov 1st 2012, 06:19 PM
Eraser147
Natural log and smallest number
Attachment 25517 Please click image to enlarge. I don't understand why the answer becomes Ln pi/2 - 1.0.
• Nov 1st 2012, 06:30 PM
chiro
Re: Natural log and smallest number
Hey Eraser147.

Try undoing the transformations. You can apply an inverse cosine to get the possible solutions (and use the fact that cos is periodic).

So then you have e^x + 1 = n*2*pi + pi/2 fo integer values n.

Now x has to be minimized for this to hold and e^x is unique and > 0 which means that you can't have negative values for n so n = 0. This means e^x + 1 = pi/2 since pi/2 > 1 but less than all the others (the minimal branch).

So now you have e^x = pi/2 - 1. and taking logarithms gives you x = ln(pi/2 - 1).
• Nov 1st 2012, 06:30 PM
MarkFL
Re: Natural log and smallest number
The given equation implies:

$\displaystyle e^x+1.0=\frac{\pi}{2}+k\pi$

Since $\displaystyle 0<1<\frac{\pi}{2}$ we want $\displaystyle k=0$ and so:

$\displaystyle e^x+1.0=\frac{\pi}{2}$

$\displaystyle e^x=\frac{\pi}{2}-1.0$

Convert from exponential lo logarithmic form:

$\displaystyle x=\ln\left(\frac{\pi}{2}-1.0 \right)$