By applying logarithm laws and the relationship between logarithms and powers of same base, the expression [tex]\log_{2} \frac{8\cdot x^{3}}{2} = \log_{2} 8\cdot x^{3} - \log_{2} 2\cdot y[/tex] is equal to y = 1.
Herein we must simplify an expression that uses logarithms by applying any of the following three laws:
[tex]\log_{a} (b \cdot c) = \log_{a} b + \log_{a} c[/tex] (1)
[tex]\log_{a} \left(\frac{b}{c} \right) = \log_{a} b - \log_{a} c[/tex] (2)
[tex]\log_{a}{b^{c}} = c \cdot \log_{a} b[/tex] (3)
Now we proceed to simplify the expression:
[tex]\log_{2} \frac{8\cdot x^{3}}{2} = \log_{2} 8\cdot x^{3} - \log_{2} 2\cdot y[/tex]
[tex]\log_{2} 2\cdot y = \log_{2} 8\cdot x^{3} - \log_{2} \frac{8\cdot x^{3}}{2}[/tex]
[tex]\log_{2} 2 \cdot y = \log_{2} \frac{\frac{8\cdot x^{3}}{1} }{\frac{8\cdot x^{3}}{2} }[/tex]
[tex]\log_{2} 2\cdot y = \log_{2} 2[/tex]
By the relationship between logarithms and powers of same base:
2 · y = 2
y = 1
By applying logarithm laws and the relationship between logarithms and powers of same base, the expression [tex]\log_{2} \frac{8\cdot x^{3}}{2} = \log_{2} 8\cdot x^{3} - \log_{2} 2\cdot y[/tex] is equal to y = 1.
To learn more on logarithms: https://brainly.com/question/20785664
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