Plzz help gahhhhh :\
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Answer:
The product of 'x' and 'y' is [tex]\boxed 8[/tex].
Step-by-step explanation:
Given:
[tex]\log_{5\sqrt5}125=x\\\\\log_{2\sqrt2}64=y[/tex]
We need to determine the product of 'x' and 'y'.
Using the following logarithmic property:
[tex]\log_ab=\frac{\log b}{\log a}[/tex]
Here, [tex]a=5\sqrt5\ and\ 2\sqrt2[/tex]
[tex]b=125\ and\ 64[/tex]
So, [tex]log_{5\sqrt5}125=\frac{\log 125}{\log 5\sqrt{5}}\\\\log_{5\sqrt5}125=\frac{\log 5^3}{\log 5\times5^{1/2}}.......[\sqrt5=5^{1/2}][/tex]
[tex]log_{2\sqrt2}64=\frac{\log 64}{\log 2\sqrt{2}}\\\\log_{2\sqrt2}64=\frac{\log 2^6}{\log 2\times2^{1/2}}.......[\sqrt2=2^{1/2}][/tex]
Now, we use another property of log and exponents.
[tex]\log a^m=m\log a\\a^m\times a^n=a^{m+n}[/tex]
[tex]log_{5\sqrt5}125=\frac{3\log 5}{\log 5^{1+{1/2}}}=\frac{3\log 5}{\log 5^{\frac{3}{2}}}=\frac{3\log 5}{\frac{3}{2}\log 5}=2\\\\\\\\log_{2\sqrt2}64=\frac{6\log 2}{\log 2^{1+{1/2}}}=\frac{6\log 2}{\log 2^{\frac{3}{2}}}=\frac{6\log 2}{\frac{3}{2}\log 2}=\frac{12}{3}=4[/tex]
So, [tex]x=2\ and\ y=4[/tex]
The product of 'x' and 'y' = [tex]2\times 4=8[/tex]
Therefore, the product of 'x' and 'y' is 8.