How to calculate $\ln(x)$ I know only to calculate $\ln()$ using a calculator, but is there a way to calculate it without calculator:
for example: $\ln(4)= ??$ as far as I know the only way to do so is to draw the graph of $ln$ but it primarily depends on values from calculator. So really is there some formula for $\ln()$ to calculate its exact value for a specific number?
 A: You can use the following definition for the natural logarithm: $$\ln t=\int_1^t\frac{1}{x}dx.$$ So $\ln t$ represents the area below the graph of $1/x$ from $x=1$ to $x=t$. You can easily approximate this area using the Trapezium Rule or other similar approximation methods.
A: For a really efficient method, see The Arithmetic-Geometric Mean and Fast Computation of Elementary Functions. This paper showed how to obtain $\ln$ efficiently, from which all other elementary functions can be obtained with the same efficiency. Firstly $\exp$ can be obtained using Newton-Raphson approximation, but with each iteration carried out with precision being only proportional to the square of the error in the previous iteration, according to the proof I gave in another answer. (The starting value also needs to be computed carefully.) This would mean that the time taken in the last iteration is at least a constant fraction of the total time, and so computing $\exp$ has the same time complexity as $\ln$. Then using $\exp$ we can obtain $\cos$ and $\sin$ using $\cos(z) = \frac{e^{iz}+e^{-iz}}{2}$ and $\sin(z) = \frac{e^{iz}-e^{-iz}}{2i}$. Finally we can get $\tan$ from $\cos$ and $\sin$, and all their inverses using Newton-Raphson approximation again.
There is also a less efficient method that is simpler to understand, which proceeds by computing $\exp$ using argument reduction and then the identity $\exp(z) = \cos(z) + i \sin(z)$, followed by Taylor expansion for $\cos$ and $\sin$. Firstly, $\exp(z) = \exp( z \cdot 2^{-k} )^{2^k}$, and for $p$ bits of precision we can choose the minimum integer $m$ such that $z < 2^m$ and then the minimum integer $k > m + \sqrt{p}$. Then $| z \cdot 2^{-k} | < 2^{-\sqrt{p}}$, thus only $O(\sqrt{p})$ terms of the Taylor expansions are needed, which needs $O(\sqrt{p})$ multiplications at $(p+O(k))$ bits of precision. After that, the result is simply squared $k$ times, which only loses $O(k)$ bits of precision.
A: You could use it Taylor Series.
Define
$$\operatorname{ln}(y) = -\operatorname{ln}(1-x) = \sum_{k=0}^{\infty}\frac{x^k}{k}$$
Then you are able to calculate the logarith for all real numbers $y>1$ with $y=\frac{1}{1-x}$

For real numbers $0<y\leq1$ you could use
$$\operatorname{ln}(y) = (y-1)-\frac{(z-1)^2}{2} + \frac{(z-1)^3}{3} \mp \dots$$
instead
A: For $t$ near $1$, you can use the Taylor series for $\ln(1+x) $, with $x=t-1$. For other values, you use tricks. For instance, $\ln4=-\ln1/4$. As $1/4$ is kind of far from $1$, it is better to use $\ln4=2\ln2=-2\ln1/2$. 
A: There is no simple formula. There are lots of different methods, some simpler, some better, some usable if you want a million decimal digits of $\ln 4$. I'll show you what I think is the most elementary method; all it requires is that you calculate the number $e = 2.718...$ with enough precision. (However, the method works for base $10$ or base $2$ logarithms as well, and there you need no calculation to get the base). 
The only serious maths that you need is that $\ln (e) = 1$, $\ln (xy) = \ln (x) + \ln (y)$, and $\ln (x^2) = 2 \ln (x)$. 
To start the calculation of $\ln x$ for some $x > 0$, you start with a sum $s = 0$. 
As long as $x ≥ e$, divide $x$ by $e$ and increase the sum $s$ by 1. 
As long as $x < 1$, multiply $x$ by $e$ and decrease the sum $s$ by 1. 
(If $x$ was equal to $e^n$ and therefore $\ln x = n$, you will have $x = 1$ and $s = n$ at this point. If $e^n ≤ x < e^{n+1}$, you will have $s = n$, and $x$ will be divided by $e^n$, so $1 ≤ x < e$). 
Now calculate the square of $x$, that is $x = x^2$. If now $x ≥ e$, divide $x$ by $e$ and increase $s$ by $1/2$. 
Calculate the square of $x$ again, and if $x ≥ e$, divide $x$ by $e$ and increase $s$ by $1/4$. Repeat this again and again, each time adding only half as much as in the previous step, until your result is precise enough for you. 
This works because every time $x$ is squared, you double the logarithm. So to compensate for that, you only take half the logarithm of $x$, then a quarter, and so on. 
This is by far not the most efficient way, but I think it's the easiest one to explain. 
A: to find ln(a),
find n, such that e^n &lt a &lt e^(n+1)
x=a/e^n so that ln(a) = n + ln(x) and 1 &lt x &lt e
find y ,such that
x=(1+y)/(1-y)            (x>1 is assumed)
y=(x-1)/(x+1)
hence, ln(x)=ln(1+y)-ln(1-y)
            =2(y + y^3/3 + ...)
this converges rapidly as y &lt 1 always.
for x &lt 1, ln(x)=-ln(1/x) 
