swim4it52 said:
I don't understand exactly what a natural log is or how it is useful.
Playing with compound interest, the Natural Log is a "natural" result of the compounding process.
Try this experiment - compounding interest for one (1) year:
One compounding period/year: (1+i/1)<sup>1</sup>
Two compounding periods/year: (1+i/2)<sup>2</sup>
Four compounding periods/year: (1+i/4)<sup>4</sup>
Eight compounding periods/year: (1+i/8)<sup>8</sup>
Sixteen compounding periods/year: (1+i/16)<sup>16</sup>
If you pick an interest rate, and calculate these values, and then keep calculating more with the compounding periods increasing, you will notice that the value continues to increase. It increases each time, but by less and less and less. This may cause one to pause and ask, "How high will it go?" or "Is there a limit?" The answer to these question is what leads to the exponential, Base 'e', which is directly related to the Natural Logarithm. most find this fascinating that we weren't even playing with 'e' and out it popped, anyway. That is a major reason why it is called the base of the "Natural" Logarithm. It just keeps popping up, all by itself.
Logarithms with which you may be familiar are probably Base 10 logarithms. The represent exponents on a Base of 10. The natural Logarithm is an exponent on a Base of 'e'.
Surprise, the limit of the process shown above is e<sup>i</sup>! Who would have guessed? It is usually referred to as "Compounded Continuously".
Initial Amount = A
Compounding Periods per Year = n
Years = t
i = Nominal Annual Interest Rate
Future Amount = A*(1 + i/n)^(n*t)
The question is, Future Amount = F*A, for arbitrary F.
F*A = A*(1 + i/n)^(n*t)
Divide by the common factor, A
F = (1 + i/n)^(n*t)
Logarithms - It's the only way to get things out of exponents.
log(F) = log((1 + i/n)^(n*t)) = (n*t)*log(1 + i/n)
Note: These are just logarithms. To this point, any base will do just as well. Later, it will become convenient to use Base e.
t = log(F)/[n*log(1 + i/n)]
For a given value of n, that mess in the denominator is just a big, ugly constant, leaving t = C*log(F), which is the premise we were hoping to demonstrate.
a) F = 2, and t = 3 -- 3 = log(2)/[n*log(1 + i/n)]
b) n = 4 -- 3 = log(2)/[4*log(1 + i/4)] -- Solve that for i.
I'm not really quite sure what the questions mean, so I'll just leave those generalities. "What is the balance?" What balance? "If it is compounded quarterly..." If what is compounded quarterly?
