Question

If `n^(5/2) = 32`, what is the value of `n`?

Answer 1

`n=4`

Explanation:

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(n^(5/2)=(sqrtn)^5)color(white)(2/2)|)))`

Take the `color(blue)"fifth root of both sides"`

`rArr((sqrtn)^5)^(1/5)=root(5)(32)`

`rArrsqrtn=2`

To solve for n, `color(blue)"square both sides"`

`(sqrtn)^2=2^2rArrn=4`

Answer 2

`n=4`

Explanation:

We should write `32` using prime factorization.

Note that `32` in its most basic form is `32=2xx2xx2xx2xx2`, or `32=2^5`.

So, we have:

`n^(5/2)=2^5`

Now, we want to solve for `n`. To undo the power of `5//2`, we should raise both sides of the equation to the power of `2//5`. Note that `(n^(5/2))^(2/5)=n^(5/2xx2/5)=n^1=n`.

So, we have:

`(n^(5/2))^(2/5)=(2^5)^(2/5)`

`n=2^(5xx2/5)=2^2=4`

Answer 3

The Real solution is `n = 4`

There are additional Complex solutions:

`n = 4cos((4pi)/5)+4isin((4pi)/5)`

`n = 4cos(-(4pi)/5)+4isin(-(4pi)/5)`

Explanation:

The other solutions have in one way or another assumed:

`n^(ab) = (n^a)^b`

This does hold when `n` is Real and non-negative, or if `a, b` are integers, but does not hold in general.

Since the questions asks about a fractional power `5/2`, it is not safe to assume the above unless we also have some conditions on `n` - say that it is a non-negative Real number.

If `n` is non-negative Real then:

`n = n^1 = n^(5/2*2/5) = (n^(5/2))^(2/5) = 32^(2/5) = (2^5)^(2/5) = 2^(5*2/5) = 2^2 = 4`

Are there other solutions?

If `n` is allowed to take Complex values, then there are two other solutions, namely:

`n = 4cos((4pi)/5)+4isin((4pi)/5)`

and:

`n = 4cos(-(4pi)/5)+4isin(-(4pi)/5)`

To see why, consider de Moivre's formula:

`(cos theta + i sin theta)^N = cos N theta + i sin N theta`

So we find:

`(4cos((4pi)/5)+4isin((4pi)/5))^(5/2) = (4(cos((4pi)/5)+isin((4pi)/5)))^(5/2)`

`color(white)((4cos((4pi)/5)+4isin((4pi)/5))^(5/2)) = 4^(5/2)(cos((4pi)/5)+isin((4pi)/5))^(5/2)`

`color(white)((4cos((4pi)/5)+4isin((4pi)/5))^(5/2)) = 4^(5/2)(cos((4pi)/5*5/2)+isin((4pi)/5*5/2))`

`color(white)((4cos((4pi)/5)+4isin((4pi)/5))^(5/2)) = 4^(5/2)(cos(2pi)+isin(2pi))`

`color(white)((4cos((4pi)/5)+4isin((4pi)/5))^(5/2)) = 4^(5/2)`

`color(white)((4cos((4pi)/5)+4isin((4pi)/5))^(5/2)) = 32`

Similarly:

`(4cos(-(4pi)/5)+4isin(-(4pi)/5))^(5/2) = 32`