• zmudz
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 \(p + q\).
  • Stacey Warren - Expert
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  • schrodinger
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  • anonymous
\(49\) ?
  • anonymous
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 \(p+q=49\).

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