Answer:
z = 5
Step-by-step explanation:
[tex]\frac{4x^2y^7}{8x^3y^2} = \frac{y^z}{2x}[/tex]
It can be also written as -
[tex]\frac{2^2 \times x^2 \times y^7}{2^3 \times x^3 \times y^2} = \frac{1 \times x^0 \times y^z}{2 \times x^1 \times y^0}[/tex]
Using law of exponents , lets solve this.
[tex]=> 2^{(2-3)} \times x^{(2-3)} \times y^{(7 - 2)} = \frac{1}{2} \times x^{(0-1)} \times y^{(z-0)}[/tex]
[tex]=> 2^{-1} \times x^{-1} \times y^{5} = 2^{-1} \times x^{-1} \times y^z[/tex]
[tex]=> \frac{y^5}{2x} = \frac{y^z}{2x}[/tex]
[tex]=> y^z = y^5[/tex]
[tex]=> z = 5[/tex]
