how to calculate 2^1.4 So I have got a very basic question but it didn't come up as a google search so I am posting it here.

I want to know how to easy calculate 

2^1.4 = 2.6390...

Using log and antilogs i.e not easy approach ?

i.e.

log y =  log m^n

log y = n log m

log y = 1.4 log 2

log y = 1.4 * 0.301

log y = 0.4214

antilog (log y) = antilog 0.4214

y= antilog 0.4212 (look for this in a table, should give result)

What I found on internet ,

turn the decimal into a fraction 

2^(1.4) 
=> 2^(1 + 2/5) 

now given a^(n + m) = a^n * a^m 
=> 2^1 * 2^(2/5) 
=> 2 * 2^(2/5) 

you can stop there if you want to. 

however a^(nm) = (a^n)^m 

=> 2 * (2^2)^(1/5) 


  //I was ok till here what to do next to get the right result !


which would be read as 2 times the 5th root of 2 squared and would be the exact answer. [the nth root of a = a^(1/n)]
You can use Newtons approximation method. 

We want to find $x=2^{1.4}$, or equivalently, $x^5=\left(2^{1.4}\right)^5=2^7=128$

Define $f(x)=x^5-128$

We want to find the root of $f(x)$

as noted in my comment, $x\approx \sqrt{8} \approx 2.828$

So we start with this guess of $2.828$.

We put it into the formula,

$$x'=x-\frac{f(x)}{f'(x)}$$

and $x'$ will be a more accurate guess.

$$x'= 2.828 - \frac{(2.828)^5-128}{5(2.828)^4}$$

with a bit of hand calculation we get $x'=2.66\ldots$ which is very close to the actual value.
2^1.4 =(1+1)^1.4   approximately 
           =1+1×1.4     approximately 
           =1+1.4         approximately 
           =2.4             approximately 
For the above I used binomial theorem