Logarithms - how long does it take to grow? Understanding the example Here is the quote from the article I am reading on logs:

...interpretations of the natural logarithm (ln(x)), i.e. the natural log of 1.5:
Assuming 100% growth, how long do you need to grow to get to 1.5? (.405, less than half the time period)
Assuming 1 unit of time, how fast do you need to grow to get to 1.5? (40.5% per year, continuously compounded)
Logarithms are how we figure out how fast we're growing. 
Link to the original post
I don't understand what he is referring to here. 
For example, if to base an assumption of what he is saying, then if I want to know how long it will take me to get 100, I have to 
$$ln(100) = 4.605$$
So, I would say (I might be very wrong, that is why I am here asking for help) that the formula above means:
I have some initial "period" (or what else is it?) which equals 2.718 (of what?), and to get to 100 I need to:
$$2.718 ^4.605$$
I don't understand that logic and the meaning of this endeavor - what does the epsilon period gives us? 
I will be grateful for your help in finding my mistake and helping me to understand the meaning. 
Thank you very much!
 A: I am not convinced that your link explains this at all well. Let's instead take an example based on your figures:

How many years of $50\%$ interest compounded would it take to turn $\$1$ to $\$100$?

You could write this as having to solve $1 \times 1.5^n = 100$ 
Taking logarithms would change this to $\ln(1) +n \ln(1.5) = \ln(100)$ 
and we know $\ln(1)=0, \ln(1.5)\approx 0.4054651, \ln(100)\approx 4.60517$ 
so we get $n \approx \frac{4.60517}{0.4054651} \approx 11.36$ years. This makes sense as $1.5^{11} \approx 86.5$ and $1.5^{12} \approx 129.75$
You have done that calculation with logarithms to any base.  
One of the  attractive features of natural logarithms base $e$ is that, for $x$ close to $0$, you have $\ln(1+x)\approx x$, related to $\left(1+\frac1n\right)^n\approx e$ for large $n$.  So suppose in the previous question the interest rate had been a much smaller $0.05\%$ compounded, we would have found $\ln(1.0005) \approx 0.0005$ (actually nearer $0.000499875$) and so we might approximate the number of years to reach $100$ with $\frac{4.60517}{0.0005} \approx 9210$ ($9212.6$ is closer to the true result, but this is close enough for an illustration)
