In our finance course , we are allowed to use Simple Calculators that do not have a log function in them. We are provided with Log Tables but we are taught this trick to solve log without using log tables. I am not sure how this works.
Say for example I want to compute the Future value of Rs 100 invested at the rate of 12.31% p.a. compounded continuously. So the formula for continuous compounding is $$A= P.e^{rn} $$
So as a work arround we use the following method: We solve for $e^{rn}$
Let $x = e^{rn}$
Taking Log on both sides
$ Log x = r.n. Log . e $
r=0.1231 and n= 1 year
$ Log x = 0.1231* 1 * Log 2.71828$
$Log x = 0.1231 * Log 2.71828$
Now on Calculator we type 2.71828 and then we press Square Root symbol 15 times, we get 1.00003051801 {upto 12 digits} We subtract 1 from it we get 0.00003051801 We divide 0.00003051801 by 0.000070274(This is the constant that we multiply every time to find log) $ {(1.00003051801-1)\over 0.000070274} = 0.43427170788 $ Now we substitute this figure in above Equation
$ Log x = 0.1231 * Log 2.71828$
$ Log x = 0.1231 * 0.43427170788$
$ Log x = 0.05345884724 $
Now we take antilog on both sides
$ Antilog (log x)= Antilog (0.05345884724)$
Similary to how we found $Log 2.71828$ we perform the whole series of steps backwards
- We first muliply the number with 0.000070274
- Then we add back one
- Then we sqaure it 15 times
${{{{{{{{{{{{{{{x^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}^2}$
and voila, we get the answer $x = 1.13099878536$
We substitue x in original Equation
$ A = P.e^{rn} $
$ A = P.x$
$ A = 100 * 1.13099878536$
$ A = 113.099878536$
Even if we ignore that step of multiplying or dividing by "0.000070274" and do sqaure root or sqaure for 12 times rather than 15 times, we still get the same answer.
Can you please explain me how this is working?
You are using the fact that $\lim_{n \to \infty}\left(1+\frac an\right)^n=e^a$ and the fact that $\ln(1+y) \approx y$ for $y \ll 1$.
When you type in your approximation for $e$ and take the square root $15$ times, you are taking the $2^{15th}$ root of $e$, which is about $1.000030518$ as you say. Subtracting $1$ gives you $\ln (e^{2^{-15}})$, which is $2^{-{15}}=\frac 1{32768}\approx 0.000305176.$
The division by $0.00070274$ is immediately reversed by the multiplication, so you may skip both of these steps.
Multiplying by $r$ and adding $1$ gives you $1+\frac r{32768}$ and squaring $15$ times gives $\left(1+\frac r{32768}\right)^{32768}\approx e^r$
It would be far easier to just divide $r$ by $32768$, add $1$, and square $15$ times. You could also divide by $2^{12}=4096$, add $1$ and square $12$ times. The first is more accurate because you are using a larger $n$. It is less accurate because you lose one more digit of $1+\frac rn$ to the fixed precision of your calculator. Particularly if $r$ is small, the second will dominate and smaller $n$ will be better.