how to calculate 2^1.4
By Matthew Cannon
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)]
$\endgroup$ 85 Answers
$\begingroup$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.
$\endgroup$ 7 $\begingroup$$$2^{1.4}=e^{1.4\ln 2}=\sum_{k=0}^\infty\frac{(1.4\ln 2)^k}{k!}$$
$\endgroup$ $\begingroup$I don't know if you're allowed to use that Log(2) = .693147... or not. If you are you can do the following:
2^1.4 = 2 2^.4 = 2 2^.5 2^-.1 .
Now you need only approximate 2^-.1 . 2^-x for small x: 2^-x = 1 - Log[2] x + 1/2 Log[2]^2 x^2 - 1/6 Log[2]^3 x^3 + ...
Now set x=0.1 and keep as many terms as needed for your desired accuracy. You get 2^x = 1, .9307..., .933088..., .933032..., .933033.... . The last of these (.933033) is accurate to 6 decimal places. Thus your answers are
2 Sqrt[2] 2^-.1 = 2.82843... (just keeping the leading term of 1), 2.63238... (keeping x), 2.63917... (keeping x & x^2), 2.63901... (keeping up to 3rd order), 2.63902... (up to 4th order), 2.63902... (up to 50th order)
$\endgroup$ $\begingroup$If it is a multiple choice question then you can also try the possible answers using Sabyasachi's equation: if $x$ is such that $2^{1.4}=x$ then $x^5=128$. Therefore, if the answers are $2.63,2.64$ and $2.65$ then calculate $2,64^5$. You will see that it is larger then $128$ so you also want to try $2,63^5$ and choose whichever is closer to 128.
$\endgroup$ $\begingroup$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
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