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

\(49\) ?

Answer 2

It's kind of a guess, but there is an explanation behind it. I'll leave it to more meticulous minds to check if I made an error somewhere...

After some playing around, I thought to do something like this: Consider two of the equations taken at a time (so that you look at three equations).
\[\begin{cases}
~~~\color{red}{2\log_x(2y)}=\color{blue}{2\log_{2x}(2^2z)}&&&(1)\\
~~~\color{red}{2\log_x(2y)}=\color{green}{\log_{2x^4}(2^3yz)}&&&(2)\\
\color{blue}{2\log_{2x}(2^2z)}=\color{green}{\log_{2x^4}(2^3yz)}&&&(3)
\end{cases}\]
Rearrange \((1)\):
\[\begin{align*}
\color{red}{2\log_x(2y)}&=\color{blue}{2\log_{2x}(2^2z)}\\[1ex]
\frac{\ln(2y)}{\ln x}&=\frac{\ln(2^2z)}{\ln(2x)}\\[1ex]
\frac{\ln(2x)}{\ln x}&=\frac{\ln(2^2z)}{\ln(2y)}\\[1ex]
\frac{\ln2+\ln x}{\ln x}&=\\[1ex]
1+\log_x2&=\log_{2y}(2^2z)\\[1ex]
1&=\log_{2y}(2^2z)-\log_x2&(4)
\end{align*}\]
Rearrange \((2)\):
\[\begin{align*}
\color{red}{2\log_x(2y)}&=\color{green}{\log_{2x^4}(2^3yz)}\\[1ex]
\frac{\ln(2y)}{\ln x}&=\frac{\ln(2^3yz)}{2\ln(2x^4)}\\[1ex]
\frac{\ln(2x^4)}{\ln x}&=\frac{\ln(2^3yz)}{2\ln(2y)}\\[1ex]
\frac{\ln2+4\ln x}{\ln x}&=\\[1ex]
1+\frac{\ln2+3\ln x}{\ln x}&=\\[1ex]
1+\log_x(2x^3)&=\\[1ex]
1&=\frac{1}{2}\log_{2y}(2^3yz)-\log_x(2x^3)&(5)
\end{align*}\]
Set the RHS's of \((4)\) and \((5)\) to be equal, so you have
\[\begin{align*}
\log_{2y}(2^2z)-\log_x2&=\frac{1}{2}\log_{2y}(2^3yz)-\log_x(2x^3)\\[1ex]
\log_{2y}\left(\frac{2^2z}{\left(2^3yz\right)^{1/2}}\right)&=\log_x\left(\frac{2}{2x^3}\right)\\[1ex]
\frac{1}{2}\log_{2y}\frac{2z}{y}&=-3\\[1ex]
(2y)^{-6}&=\frac{2z}{y}\\[1ex]
\frac{1}{2^7}&=y^5z&(6)
\end{align*}\]
This result is super-convenient, now you need only find \(x\).

Rearrange \((3)\):
\[\begin{align*}
\color{blue}{2\log_{2x}(2^2z)}&=\color{green}{\log_{2x^4}(2^3yz)}\\[1ex]
\frac{\ln(2^2z)}{\ln(2x)}&=\frac{\ln(2^3yz)}{2\ln(2x^4)}\\[1ex]
\frac{\ln(2x^4)}{\ln(2x)}&=\frac{\ln(2^3yz)}{2\ln(2^2z)}\\[1ex]
\frac{\ln(2x)+3\ln x}{\ln(2x)}&=\\[1ex]
1+3\log_{2x}x&=\\[1ex]
1&=\frac{1}{2}\log_{2^2z}(2^3yz)-3\log_{2x}x&(7)
\end{align*}\]
Substitute for \(z\) by using \((6)\):
\[\large
\begin{align*}
1&=\frac{1}{2}\log_{2^2\left(2^{-7}y^{-5}\right)}\left(2^3y\left(2^{-7}y^{-5}\right)\right)-3\log_{2x}x\\[1ex]
&=\frac{1}{2}\log_{2^{-5}y^{-5}}\left(2^{-4}y^{-4}\right)-3\log_{2x}x\\[1ex]
&=-2\log_{2^{-5}y^{-5}}(2y)-3\log_{2x}x\\[1ex]
&=-2\frac{\ln(2y)}{-5\ln(2y)}-3\log_{2x}x\\[1ex]
&=\frac{2}{5}-3\log_{2x}x\\[1ex]
\frac{3}{5}&=-3\log_{2x}x\\[1ex]
(2x)^{-1/5}&=x\\[1ex]
x^6&=\frac{1}{2}\\[1ex]
x&=\frac{1}{2^{1/6}}
\end{align*}\]
(ignoring the negative root, since \(x,y,z>0\)).

So, you have
\[xy^5z=\frac{1}{2^{1/6}}\times\frac{1}{2^7}=\frac{1}{2^{43/6}}\]
i.e. \(p=43\) and \(q=6\), which are relatively prime. Therefore \(p+q=49\).