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zmudz
 one year ago
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 \(p + q\).
zmudz
 one year ago
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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Loser66
 one year ago
Best ResponseYou've already chosen the best response.0Just curious :) Which class is it?

freckles
 one year ago
Best ResponseYou've already chosen the best response.1well 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: \[u=\log_x(4y^2) \\ x^u=4y^2 \] \[u=\log_{2x}(16z^2) \\ x^{u}=2^{4u}z^2 \\\] \[u=\log_{2x^4}(8yz)=u \\ x^{u}=2^{\frac{3u}{4}}(yz)^\frac{1}{4}\] so I setup a system of equations \[2^2y^2=2^{4u}z^2 \\ 2^{4u}z^2=2^{3u}{4}(yz)^\frac{1}{4} \\ 2^2y^2=2^{\frac{2u}{4}}(yz)^\frac{1}{4} \\\] Eventually got: \[z^2=2^{2+u}y^2 \text{ then also } \\ y^2=2^\frac{12u}{6} \text{ and then found } z=2^{\frac{5u24}{6}}\] \[2^{4u}2^{5u24}{6}=2^\frac{3u}{4}2^\frac{12u}{48}2^\frac{5u24}{48} \\ \text{ use law of exponents to get the equation } \\ 2^{4u+\frac{5u24}{6}}=2^{\frac{3u}{4}+\frac{12u}{48}+\frac{5u24}{48}} \\ \text{ set the exponents on both sides equal \to each other } \\ \text{ you can find } u \text{ now }\] I plug this u into \[z^2=2^{2+u}y^2 \text{ then also } \\ y^2=2^\frac{12u}{6} \text{ and then found } z=2^{\frac{5u24}{6}}\] 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 \[x^{u}=4y^2\]

freckles
 one year ago
Best ResponseYou've already chosen the best response.1Don't know if there is a shorter way.
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