Solve. 4^(3/5) without calculator?

Question

I can only solve for 2 as denominator eg. y^(x/2) because it would be (2 root y)^x and is simple.

I know that the problem above would be 3 root 4 to the 5th power, but how would you solve from there without a calculator? or can you?

Thanks!

Answer 1

As others have shown, the answer is 2 * 2^(1/5),

which is approximately 2.2974.

This can be solved without a calculator using Newton's

iterative method, although with each iteration, the

calculations soon become a life's work. How much

work you do depends on the accuracy you need.

Here are two iterations :

We want to solve x = 2^(1/5) and later, multiply by 2.

So, x^5 = 2 and x^5 - 2 = 0.

Let f(x) = x^5 - 2. Therefore, f '(x) = 5x^4.

The formula is x(n) = x(n-1) - {f[x(n-1)] - Q} / f '[x(n-1)]

where Q = 2 and n and n-1 are actually subscripts.

We'll start with n = 1 and call our first guess x(0) = 1.

So, f[x(0)] = f(1) = 1 - 2 = -1, and f '[x(0)] = f '(1) = 5.

Thus, x(1) = 1 - (-1)/5 = 1.2, so x = 2 * 1.2 = 2.4.

Now, x(2) = 1.2 - [(1.2)^5 - 2] / [5*(1.2)^4]

which is a bit more difficult, but possible in a short time.

Just multiply 1.2 by itself 5 times, then subtract 2.

Then divide by 5, then divide by 1.2, four times, then

subtract it from 1.2.

Less than half an hours work!

The answer is 1.15, so x = 2 * 1.15 = 2.30,

which is already good to 2 decimal places,

although you should do another iteration to check,

but you'll need more than half an hour to calculate

1.15 - [(1.15)^5 - 2] / [5*(1.15)^4], I imagine.

Another way, is to make use of the binomial formula :

(x + y)^n = x^n + nx^(n-1)y + [n(n-1)/2!]x^(n-2)y^2 + ...

With x = 1 and y = 1 and n = 1/5, we have :

2^(1/5) = 1 + (1/5) + (1/5)(-4/5)/2! + (1/5)(-4/5)(9/5)/3! + ...

but unfortunately, this does not converge very quickly.

Answer 2

Try using logarithms maybe?

Eg: log4(x)=3/5

That's my best guess, play around a bit...

Answer 3

4^(3/5)

= (2^2)^(3/5)

= 2^(6/5)

= 2 x (fifth root of 2)

~ 2 (1.15) = 2.3

Answer 4

Hey,

4^(3/5) =

(4^3) = 48

5th root of 48 is some irrational number that you may able to get if you are really really really good with numbers, other wise calculator is the way to go.

hope this helps

Answer 5

Maybe there is something in your book about

x^y = x*x*x*x* ...*x until there are y of those little x guys.

Also maybe

x^(1/y) is the yth root of x: sort of the opposite of the above.

also x^(3/5) should be (x^3)^1/5 or something like that

We also know that 4 = 2*2

so 4^(3) = 4 * 4 * 4

= 2*2 * 2*2 * 2*2

so you need (2 * 2 * 2 * 2 * 2 * 2)^(1/5) which does not look that obvious.  Sure the problem was not 4^(3/6)? :)

So I guess (2*2*2*2*2)^(1/5) would be easy = 2

So maybe you want

4^3 = 2 * 2^5

4^(3/5) = (2*2*2*2*2)^(1/5) * 2^(1/5)

= 2 * 2^(1/5)

= 2 * 5root(2) ?

Maybe someone can see a different step?

Answer 6

4^(3/5)

= ⁵√(4³)

= ⁵√(4·4·4)

= ⁵√(64)

= ⁵√(32·2)

= ⁵√(32)·⁵√(2)

= ⁵√(2⁵)·⁵√(2)

= 2·⁵√(2) = 2·2^(1/5)

Answer 7

4^(3/5) = (4^3)/(4^5)  which simplifies to 1/(4^2) or 1/16

I'm not 100% sure about this answer though.

Answer 8

4^(3/5)

= (2^2)^(3/5)

= 2^(6/5)

= 2 * 2^(1/5).