How to find the value of $x$ in $x^5=32$

Question

I understand that $2^5=32$ but how would one go about finding it without doing any guessing (what if the numbers were much greater)?

Answer 1

USING COMPLEX ANALYSIS

We have

$$z^5=2^5\implies z=2e^{i2n\pi/5}$$

for $n=0, \pm 1, \pm 2$.

$$\bbox[5px,border:2px solid #C0A000]{\text{Thus the five roots are} \,\, 2,\, 2e^{\pm i2\pi/5},\,2e^{\pm i4\pi/5}}\tag 1$$

---

FACTORORING A POLYNOMIAL

We have $x^5=32\implies x^5-2^5=0$.

We can then factor $x-2$ from the left-hand side and write

$$x^5-2^5=(x-2)(x^4+2x^3+4x^2+8x+16)=0 \tag 1$$

The quartic expression in $(1)$ can be factored into the product of two quadratics as

$$x^4+2x^3+4x^2+8x+16=(x^2+ax+4)(x^2+bx+4) \tag 2$$

where by matching coefficients in $(2)$, we find $a+b=2$ and $ab=-4$.

Solving for $a$ and $b$ reveals that $a=1+\sqrt{5}$ and $b=1-\sqrt{5}$ whence the original polynomial can be written as

$$x^5-2^5=(x-2)(x^2+(1+\sqrt{5})x+4)(x^2+(1-\sqrt{5})x+4) \tag 3$$

Finally, the quadratic terms in $(3)$ can easily be factored as

$$x^2+(1+ \sqrt{5})x+4=\left(x-\frac{(1+\sqrt{5})+i\sqrt{10-2\sqrt{5}}}{2}\right)\left(x-\frac{(1+\sqrt{5})-i\sqrt{10-2\sqrt{5}}}{2}\right)$$

and

$$x^2+(1- \sqrt{5})x+4=\left(x-\frac{(1-\sqrt{5})+i\sqrt{10+2\sqrt{5}}}{2}\right)\left(x-\frac{(1-\sqrt{5})-\sqrt{10+2\sqrt{5}}}{2}\right)$$

Thus the five roots of $x^5-2^5=0$ are

$$\bbox[5px,border:2px solid #C0A000]{2,\, \frac{(1+ \sqrt{5})\pm i\sqrt{10- 2\sqrt{5}}}{2},\,\frac{(1- \sqrt{5})\pm i\sqrt{10+ 2\sqrt{5}}}{2}} \tag 4$$

where the roots in $(4)$ are the rectangular coordinate form of the polar roots in $(1)$.

Answer 2

The other answers are great and comprehensive, but from the phrasing of your question I suspect they might overwhelm you at this stage. This is hardly a real answer, but I'm guessing it's more along the lines of what you're looking for.

You have to find the $5^\text{th}$ root of $32$. Just like if you have $x^2 = y$ you take the square root of $y$ to get $x$, if you have $x^n = y$, you take the $n^\text{th}$ root of $y$ to get $x$. This corresponds to raising $y$ to the power of $1/n$. In Google you can type in "32 ^ (1 / 5)" to solve your particular equation. A scientific calculator will also allow you to compute arbitrary powers.

For now, it probably suffices to say that considering the 'normal' numbers you are familiar with, the equation $x^n = y$ (where $n$ is a positive whole number and $y>0$) has one solution if $n$ is odd, and two if $n$ is even. You start by finding $y^\frac{1}{n}$. If $n$ is odd, that is the only solution. If $n$ is even, the negative of $y^\frac{1}{n}$ is also a solution. In the example above, $n=5$ was odd, so there was only one solution. But as another example, take $x^4 = 16$. Then, computing $16^\frac{1}{4}$ gives you $x=2$, but note that $x=-2$ is also a solution because $(-2)^4 = 16$.

In the future, you might learn about complex numbers, which would show you that the equation $x^n = y$ (where $n$ is a positive whole number and $y$ is any number) has, in general, $n$ solutions.

Answer 3

Given a positive real number $x$ the $n$th root of $x$ is defined as $y=\sqrt[n]{x}$ such that $y^n=x$. we can also use a rational exponent to indicate such $n$th root: $\sqrt[n]{x}=x^{\frac{1}{n}}$.

So the solution of the equation $x^n=a$ is simply written $x=\sqrt[n]{a}=a^{\frac{1}{n}}$.

If you want explicitly evaluate the root when it is not immediate, you can use logarithms, using the fact that $ \log x=\dfrac{\log a}{n}$. Years ago (before computers) this was the standard way to do such computations, using the table of logarithms (usually in base $10$) so that , whan $\log_{10} x$ is determined from the tables, we can find $x=10^{\log_{10} x}$.

There is also a relatively easy and fast algorithm to calculate the $n$th root as you can see here.

Answer 4

If the powers are much greater, you can resort to numerical methods such as Newton's Method. For example, if we want to compute $x = \sqrt[5]a$, we need to rewrite the equation as $f(x) = x^5 - a$. Then the Newton step formula ends up being $$\begin{align}x_{n+1} & = x_n - \frac{f(x_n)}{f'(x_n)}\\ & = x_n - \frac{x^5_n - a}{5x_n^4} \\ & = \frac{a + 4x^5_n}{5x^4_n}\end{align}$$

Then, with any initial guess $x_0$ (except $x_0 = 0$ for obvious reasons), iterating the formula for many iterations will converge to the answer. So for example, if we want to compute $x = \sqrt[5]{28629151}$, then let's say our initial guess is $x_0 = 40$. Then,

$$x_1 = \frac{28629151 + 4(40)^5}{5(40)^4} \approx 34$$ Use this in the iteration formula... $$x_2 = \frac{28629151 + 4(34)^5}{5(34)^4} \approx 31.5$$ do this a third time... $$x_3 = \frac{28629151 + 4(31.5)^5}{5(31.5)^4} \approx 31.01$$ and again... $$x_4 = \frac{28629151 + 4(31.01)^5}{5(31.01)^4} \approx 31.000006$$ So it looks like this is closing in on 31. And it just so happens that $31^5 = 28629151$. So yes, while it requires some guessing, ANY nonzero guess you start off with will eventually hit the correct answer. This means you could be flat out WAAAY off and still get the right answer if you just keep iterating the formula using the previous answer you get.

If we try this on $x = \sqrt[5]{32}$, then the following happens. Maybe I think the fifth root of 32 is 5. So $x_0 = 5$ and $a = 32$. Then $$x_1 = \frac{32 + 4(5)^5}{5(5)^4} \approx 4$$ $$x_2 = \frac{32 + 4(4)^5}{5(4)^4} \approx 3$$ $$x_3 = \frac{32 + 4(3)^5}{5(3)^4} \approx 2.5$$ $$x_4 = \frac{32 + 4(2.5)^5}{5(2.5)^4} \approx 2.1$$ $$x_5 = \frac{32 + 4(2.1)^5}{5(2.1)^4} \approx 2.009$$ $$x_6 = \frac{32 + 4(2.009)^5}{5(2.009)^4} \approx 2.00008$$

So it seems like the answer should be 2. And that's what we want.