Let \(x, y,\) and \(z\) be positive real numbers that satisfy \(2 \log_x (2y) = 2 \log_{2x} (4z) = \log_{2x^4} (8yz) \neq 0.\) The value of \(xy^5 z\) can be expressed in the form \(\frac{1}{2^{p/q}}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p + q\).

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Let \(x, y,\) and \(z\) be positive real numbers that satisfy \(2 \log_x (2y) = 2 \log_{2x} (4z) = \log_{2x^4} (8yz) \neq 0.\) The value of \(xy^5 z\) can be expressed in the form \(\frac{1}{2^{p/q}}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p + q\).

Mathematics
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Just curious :) Which class is it?
precalc
serious??? no way!!

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well what I did was set each one of those things to u and then I solved for u and then went back and solved for y then z finally went back and found x. then plugged into xy^5z and did get something in the form of 2^(-p/q) then I just did p+q so first step: \[u=\log_x(4y^2) \\ x^u=4y^2 \] \[u=\log_{2x}(16z^2) \\ x^{u}=2^{4-u}z^2 \\\] \[u=\log_{2x^4}(8yz)=u \\ x^{u}=2^{\frac{3-u}{4}}(yz)^\frac{1}{4}\] so I setup a system of equations \[2^2y^2=2^{4-u}z^2 \\ 2^{4-u}z^2=2^{3-u}{4}(yz)^\frac{1}{4} \\ 2^2y^2=2^{\frac{2-u}{4}}(yz)^\frac{1}{4} \\\] Eventually got: \[z^2=2^{-2+u}y^2 \text{ then also } \\ y^2=2^\frac{-12-u}{6} \text{ and then found } z=2^{\frac{5u-24}{6}}\] \[2^{4-u}2^{5u-24}{6}=2^\frac{3-u}{4}2^\frac{-12-u}{48}2^\frac{5u-24}{48} \\ \text{ use law of exponents to get the equation } \\ 2^{4-u+\frac{5u-24}{6}}=2^{\frac{3-u}{4}+\frac{-12-u}{48}+\frac{5u-24}{48}} \\ \text{ set the exponents on both sides equal \to each other } \\ \text{ you can find } u \text{ now }\] I plug this u into \[z^2=2^{-2+u}y^2 \text{ then also } \\ y^2=2^\frac{-12-u}{6} \text{ and then found } z=2^{\frac{5u-24}{6}}\] to find y and z. Once I obtained u,y,z... I found x by going back to one of the equations I started with like \[x^{u}=4y^2\]
Don't know if there is a shorter way.

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