A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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 \(p + q\).

  • This Question is Closed
  1. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Just curious :) Which class is it?

  2. zmudz
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    precalc

  3. Loser66
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    serious??? no way!!

  4. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  5. freckles
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Don't know if there is a shorter way.

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

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.