# Math Help - Natural log and smallest number

1. ## Natural log and smallest number

Please click image to enlarge. I don't understand why the answer becomes Ln pi/2 - 1.0.

2. ## 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).

3. ## Re: Natural log and smallest number

The given equation implies:

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

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

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

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

Convert from exponential lo logarithmic form:

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