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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)]

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  • $\begingroup$ Yes, $2\left(\sqrt[5]{2}\right)^2$ is a right answer. What do you want? Decimal approximation? $\endgroup$ – Guy Apr 5 '14 at 12:10
  • $\begingroup$ Yes , I have such options 2.6390 , 2.xx $\endgroup$ – Prateek Apr 5 '14 at 12:11
  • $\begingroup$ Are you allowed to use $\log$ tables? $\endgroup$ – Guy Apr 5 '14 at 12:13
  • $\begingroup$ No as I have mentioned I want the easiest way no logs no scientific calculator ! $\endgroup$ – Prateek Apr 5 '14 at 12:13
  • $\begingroup$ Also you can note that, $$2^{1.4}\approx 2^{1.5} = \sqrt{8} \approx 2.828$$ $\endgroup$ – Guy Apr 5 '14 at 12:14
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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.

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  • $\begingroup$ This sounds easy! But the second digit after decimal is not close enough i.e. 2.66 is way different from 2.63 if there are two options 2.64 , 2.65 , 2.66 . :/ $\endgroup$ – Prateek Apr 5 '14 at 12:28
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    $\begingroup$ you can use more iterations of Newton's method to get better approximations $\endgroup$ – Alessandro Codenotti Apr 5 '14 at 12:31
  • $\begingroup$ @prateek you can repeat the same process with the new guess of $2.66$ :) Follow the link I posted. It explains more. $\endgroup$ – Guy Apr 5 '14 at 12:32
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    $\begingroup$ oh that's cool ;) $\endgroup$ – Prateek Apr 5 '14 at 12:53
  • $\begingroup$ @Sabyasachi. Your answer if fine as long as you can rationalize the exponent. What about $2^\pi$ ? I am just pushing to the limits. Cheers. $\endgroup$ – Claude Leibovici Apr 5 '14 at 13:02
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$$2^{1.4}=e^{1.4\ln 2}=\sum_{k=0}^\infty\frac{(1.4\ln 2)^k}{k!}$$

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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)

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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.

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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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  • $\begingroup$ Is this an answer? $\endgroup$ – Taroccoesbrocco Aug 18 at 13:37

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