How to manually calculate $e^x$ values? I have to appear in competitive exam, that doesn't allows calculators.
So I need to calculate exponential values for different values of 'X'.
Is there any trick or standard ways which can help us calculate e^x quiet easily ?
 A: It depends on what value you want to compute. Say you want to compute $e^x$ for small $x$. Then, as already mentioned, use the Taylor series
$$e^x = 1+x+x^2/2!+x^3/3!+\cdots.$$
But what if $x$ is large? Say it is an integer $n$. Then I would take an approximation of $e$ and compute powers of 2:
$$e, e^2, e^4, e^8, \dots, e^{2^m}.$$
The number $n$ can be expressed in binary in the form $n=b_m \cdots b_1$. So
$$e^n = e^{2^m b_m + \cdots + b_1} = \prod_{k=1}^m e^{2^k b_k}.$$
That is, you multiply all the powers of two of $e$ that appear in the binary expansion. You will have to do around $2\log_2 n$ multiplications here. I believe the relative error with not increase compared to what it was for $e$ initially. It might even decrease. If you have $x=n+\{x\}$, where $\{x\}$ is the fractional part of $x$, then you can compute $e^{\{x\}}$ with taylor series and multiple $e^n$ by that number. 
A: You can use the definition of $\exp(x)=1+x+x^2/2!+x^3/3!+\ldots$. 
Also, you can use facts like
$\exp(x+y)=\exp(x)\exp(y)$, whence, e.g., $\exp(5)=\exp(1)^5$.
A: You can also use the approximation $$\lim_{n\to \infty}\left(1+\frac xn\right)^n=e^x.$$
A: You can use the series expansion of $e^x$ to get a decent aproximation, but it isn't very comfortable to do by hand:
$e^x=\sum\limits_{k=0}^\infty\frac{x^k}{k!}=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+...$
for example to calculate $e^3$ you can calculate $1+3+\frac{9}{2}+\frac{3^3}{6}+\frac{3^4}{24}...$, the more terms you use the better your aproximation will be, in this case $e^3=20.0855...$ and the summations are $1,4,8.5,13,16.375,18.4,19.4125,19.84...$
However you have to calculate up to $3^7$ to get $19.84$, which isn't even such a precise aproximation of $e^3$
