Q:

Prime factor product

Let $P$ be the product of all prime factors $k$ and $P_d$ equal to the product of all exact dividers of $P$. Calculate or value of $x$, know that $P_d$ is equal to the product of all exact dividers of $P$. Calculate or value of $x$, know that $P_d$ is equal to $P ^ x$ . (Exact dividers: And that they all divide the number equally and that the division is exact).
I do not know, can not produce much, but I found the formula for calculating the product of dividers of a natural number:
where $a ^ k \cdot b ^ y \cdot c ^ z \cdot  ....  \cdot n ^ p$ prime factors of a number. Formula: $\sqrt (a ^ k \cdot b ^ y \cdot c ^ z \cdot ... \cdot n ^ p) ^ {(k + 1) (y + 1) (z + 1) .... (p + 1)}$
Formula of the book: ''Praticando aritmética, de Jose Carlos Admo Lacerda ''

A:

If you are told to do this for $N= 2^{k-1}$ this is easy.
$P$= product of prime factors:  $2^{k-1}$ has only $2$ as a prime factor.  So $P= 2$.
$P_d$ = product of all factors.  The factors of $2^{k-1}$ are: $1,2,4,8,.....,2^{k-1}$.  SO $P_d = 1*2*4*8*....*2^{k-1} = 2^0*2^1*2^2*2^3*....*2^{k-1} = 2^{0+1+2+3+....+k-1} = 2^{\frac {(k-1)k}2}$
And if $P_d = P^x$ then $P_d= 2^{\frac {(k-1)k}2}=2^x$ so $x = \frac {(k-1)k}2$.
To do this in general (a much harder problem), read on....
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Maybe do an example.  
$N = 2^4*3^3*5^2$
$P = 2*3*5$ and that will be easy for any $N$ with prime factorization with $N=\prod p_i^{a_i}$ then$P = \prod p_i$.
Now the factors of $N$ are $2^a3^b5^c$ where $0\le a\le 4$ and $0\le b \le 3$ and $0 \le c \le 2$.
Now what is $P_d$.  Well clearly $P_d = 2^{M_2}3^{M_3}5^{m_5}$ for some powers of $M_2, M_3, M_5$ but how do we calculate what $M_2, M_3, M_5$ are?
Well we have that $M_2$ is some sum of all the possible $a$s but how?  But for every fixed factor $3^b5^c$ will have the factors $2^0*3^b5^c$ and $2^1*3^b5^c$ and $2^2*3^b5^c$.... $2^5*3^b5^c$.  That is the factors $2^*3^b5^c$ where $b, c$ are fixed.  So for each factor $3^b5^c$ we will have $\prod (2^i*(3^b5^c))= 2^{\sum_{i=0}^4i}*(3^b5^c)^6$ and the power of two detrmined be that set of factors is $\sum_{i=0}^4i = \frac {4*5}2 =10$.
So the Those factors $2^i$ will contribute ${10}$ toward $M_2$.  Now $b$ can be $0.... 3$, that is $4$ options.  And $c$ can be $0...2$, that is $3$ options.  So there are $4*3=12$ possible $3^b5^c$ and each contribute $10$ toward $M_2$ so $M_2 = 10*12=120$ or $M_2 = (\sum_{i=0}^4 i)(3+1)(2+1)=\frac {4*5}2*(4)(3)$.
Doing the same for $M_3$ and $M_5$ we get $M_3 = (4+1)(2+1)\sum_{i=0}^3i = 5*3*\frac {3*4}2 = 15*6=90$ and $M_5 = (4+1)(3+1)\sum_{i=0}^2i = 5*4*3=60$ 
So $P_d = 2^{120}3^{90}5^{60}$.
Which happens to be $(2^43^35^2)^{30}=N^{30}$. (Although I do not see why we would have anticipated that $P_d = N^x$ for any integer value of $x.)
And in general:
If $N =\prod p_i^{a_i}$ then $Pd=\prod p_i^{M_{p_i}}= \prod p_i^{(\sum_{k=0}^{a_i} k)\prod_{j\ne i}(a_j +1)} = \prod p_i^{\frac {a_i(a_i + 1)}2*\prod_{j\ne i}(a_j +1)}=$
$\prod p_i^{\frac {a_i}2*p^{a_1+1}\prod_{j\ne i}(a_j +1)}=$
$\prod p_i^{\frac {a_i}2*\prod (a_j+1)}=$
$((\prod p_i^{a_i})^{\frac 12})^{\prod (a_j + 1)} =$
$\sqrt{\prod p_i^{a_i}}^{\prod (a_j + 1)} =$
$\sqrt{N}^{\prod (a_j + 1)}$
Which is your texts answer.
So $P_d = \sqrt{N}^{\prod (a_j + 1)} = N^{\frac 12\prod (a_j + 1)}=N^y$ where $y =\frac 12\prod (a_j + 1)$
But to solve $P_d = P^x$ can't be solved unless $N=\prod p_i^{h}$ where all powers are the same value.

