Question

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 \(p + q\).

Answer 1

Just curious :)
Which class is it?

Answer 2

precalc

Answer 3

serious??? no way!!

Answer 4

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
\[x^{u}=4y^2\]

Answer 5

Don't know if there is a shorter way.