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)]
Best Answer
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.