How to find the value of $x$ in $x^5=32$ 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)?
 A: 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 $(1)$.
A: 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. 
A: 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.
