zmudz
  • 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 \(p + q\).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\(49\) ?
anonymous
  • 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 \(p+q=49\).

Looking for something else?

Not the answer you are looking for? Search for more explanations.