Answer:
The product of 'x' and 'y' is
.
Explanation:
Given:
![\log_(5\sqrt5)125=x\\\\\log_(2\sqrt2)64=y](https://img.qammunity.org/2020/formulas/mathematics/high-school/i7c36hw8muphvlwyessbqw3x2aijj9mtsv.png)
We need to determine the product of 'x' and 'y'.
Using the following logarithmic property:
![\log_ab=(\log b)/(\log a)](https://img.qammunity.org/2020/formulas/mathematics/high-school/qrms4gje7zkk6zfdmkdz3rpb1t8xi6bwp8.png)
Here,
![a=5\sqrt5\ and\ 2\sqrt2](https://img.qammunity.org/2020/formulas/mathematics/high-school/l185yztvk7xkq0ifpo8gdzgxkaf149aiws.png)
![b=125\ and\ 64](https://img.qammunity.org/2020/formulas/mathematics/high-school/pcs87wy17727fm5ssaig556owq6uwpzj9x.png)
So,
![log_(5\sqrt5)125=(\log 125)/(\log 5√(5))\\\\log_(5\sqrt5)125=(\log 5^3)/(\log 5*5^(1/2)).......[\sqrt5=5^(1/2)]](https://img.qammunity.org/2020/formulas/mathematics/high-school/iapb7ufj5xqypx875i63ow2ujvdr5jge52.png)
![log_(2\sqrt2)64=(\log 64)/(\log 2√(2))\\\\log_(2\sqrt2)64=(\log 2^6)/(\log 2*2^(1/2)).......[\sqrt2=2^(1/2)]](https://img.qammunity.org/2020/formulas/mathematics/high-school/sruuwiuf85ghzpngnw0zl4afxldwaz5spb.png)
Now, we use another property of log and exponents.
![\log a^m=m\log a\\a^m* a^n=a^(m+n)](https://img.qammunity.org/2020/formulas/mathematics/high-school/wjg7yk1rzsv958i96d1fit65kjpmikjkjq.png)
![log_(5\sqrt5)125=\frac{3\log 5}{\log 5^{1+{1/2}}}=\frac{3\log 5}{\log 5^{(3)/(2)}}=(3\log 5)/((3)/(2)\log 5)=2\\\\\\\\log_(2\sqrt2)64=\frac{6\log 2}{\log 2^{1+{1/2}}}=\frac{6\log 2}{\log 2^{(3)/(2)}}=(6\log 2)/((3)/(2)\log 2)=(12)/(3)=4](https://img.qammunity.org/2020/formulas/mathematics/high-school/iubeoirvnr40l3xtby0b2708vwtubi8kaz.png)
So,
![x=2\ and\ y=4](https://img.qammunity.org/2020/formulas/mathematics/high-school/r37yzob2kolwfc92oczjti485qznnjha7q.png)
The product of 'x' and 'y' =
![2* 4=8](https://img.qammunity.org/2020/formulas/mathematics/high-school/sknyzi456psc49uerybtsjyycvsfipfpav.png)
Therefore, the product of 'x' and 'y' is 8.