Can you calculate 2^0.4 without a calculator?

Question

if so please show me how

Answer 1

If I wanted to approximate it fast - like now - I'd try a parabolic interpolation.

The following is entirely no-calculator:



We have the required three data points in the general vicinity of 2^0.4:

2^0 = 1

2^0.5 = √2 = approx 1.414 (well-known value to anyone who does math)

2^1 =2

... so do a curve fit for f(x) = ax^2 + bx + c with points [0,1], [0.5,1.414], [1,2]



a*0 + b*0 + c = 1 ... so c=1 <<<<<<<<<<<

0.25a + 0.5b + c = 1.414 ... a + 2b + 4c = 5.656 ... a + 2b = 1.656

a + b + c = 2 ... a + b = 1



a + 2b = 1.656

a + b =1

---------------

b = 0.656 ... so a = 1 - 0.656 = 0.344 <<<<<<<<<<<



So f(x) = 0.344x^2 + 0.656x + 1 ... approximation of 2^x in range 0 < x < 1

Now we want f(0.4)

f(0.4) = 0.344*0.16 + 0.656*0.4 + 1

= 0.05504 + 0.2624 + 1 < -- done by hand

= 1.31744 .. let's say 1.317 with reasonable rounding <<<<<<<<<



Actual value is 1.319507910



If you want a value to any precision, as others have said there are other methods such as Newton's method. But these are really not feasible without a calculator unless you have hours (or more) to spare. There is, in general, no quick, no-calculator, way to calculate arbitrary powers of a number, such as 2^0.4.

Answer 2

Figure 0.4 = 2/5.

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

So 2^.04 is the 5th root of 4.

Multiply 5 numbers together to get 4. The numbers should be as close to equal as you can get.

The square root of 2 is about 1.4142, so

1.4142*1.4142*1.4142*1.4142* 1= 4.

Take the average of those 5 numbers and you get 1.3316.

This system always gives a result a little high, so a good guess as to the answer is 1.32, which is correct to better than one hundredth.

You could keep going , figure 1.3316 is about 4/3 and try 5 numbers again, trying to keep them as close

together as possible

4/3*4/3*4/3* 4/3(81/256*4))=4

Take the average of the 5 numbers 4/3,4/3,4/3, 4/3(324/256 and your second guess is 1.319+, pretty close.

Again, if you want more accuracy, plug in 1..319 for the next iteration, and get the average of the first 5 terms in

1.319*1.319*1.319* 1.319 * (4( 1/1.319^4) =4





Here's another example. What's the square root of six? Multiply two nearly equal number to get six.

2*3-6, the average of 2 and 3 is 2.5, so 2.5 is our first guess.

2.5* ^(6/2.5) = 6 =2.5*2.4

The average of 2.5 and 2.4 is 2.45. Again, the answer is a little high, so I'd guess the correct result as 2.44

Checking on my calculator, the correct figure is 2.449+

Of course we could have done a second iteration and figured the square root of six as

2.45 *(6/2.45) = 6, Our next quess on the square root of six would have been

the average of 2.45 and 6/2.45 or 1/2 (2.45 + 2 110/245) or 2.448+

Answer 3

Yes and no.



You can't easily solve this without log tables or a calculator, but you can simplify it as follows:

0.4 = 2/5



Your expression is therefore equivalent to:

2^(2/5)



We can rewrite this as:

(2²)^(1/5)



2² is the same as 4:

4^(1/5)



And raising something to the 1/5 power is the same as taking the 5th root:

⁵√4

Answer 4

Sure, by iteration using newtons method



xnew = xold - (x^5-4)/(5x^4)



Start with a number somewhere near the correct answer. SInce 1^5 is 1 and 2^5 is 32, it's clearly

between 1 and 2. So start with 1.5



After one iteration xnew is 1.358025

Replace xold with 1.358025 and the next value for xnew will be 1.321632

another iteration gets you 1.319515

and then 1.319508

plugging in 1.319508 gets you 1.319508 again, so that's the answer to 6 decimal places.

Answer 5

this is the fifth root of 4

make use of (1+a)^5 = 1 + 5a + 10a^2 +10 a^3 +5a^4 +a^5



a is a little over .3

then fifth root of 4= fifth root of (4/1.3^5 *1.3^5))

Using basic long division, you can calculate the remainder on the left, and use the above n-1/5 calculation to narrow it down further.

Answer 6

What are the significant digits? Is the question [2.01]^(0.39) or [1.9]^(0.41) ? There was an algorithm for calculating the square root of a number I was taught in the fifth grade, but 1.414 is four significant figures and differs from the other answers by more than one or two percent. Why would you want more accuracy than precision? 1.31951 looks like a precise answer accurate to five SF http://www.mathsisfun.com/sphider/search.php?query=significant+digits&submit=Search&search=1

Answer 7

Nope. Do you realize that if you ever find a day when you need to calculate that problem and cant find a calculator...You would be better off knowing the art of flint napping and Arrow making?

Sticks and rocks 101

Answer 8

= 2^(2/5)

= 5th root of (2^2)

= 5th root of 4

It's theoretically possible but incredible difficult to figure this out manually.

You could do:

1.3^5 = 3.71293

1.4^5 = 5.37824

1.35^5 = 4.4840334375

1.325^5 = 4.083940361328125

Keep guessing until you get close enough. The actual answer is approximately 1.3195079107728942593740019712296

Answer 9

Nobody does manual powers or roots any more. Do you have a living grandfather who graduated from elementary or junior high school that you can ask? Two to the four tenths is the tenth root of two to the fourth which is the tenth root of 16 which is the fifth root of 4. Take if from there by mental iteration.

Answer 10

2^0.4=2^(2/5)=(2^2)^(1/5)=4^(1/5)=⁵√4

Answer 11

no