Quantcast

Got Homework?

Connect with other students for help. It's a free community.

  • across
    MIT Grad Student
    Online now
  • laura*
    Helped 1,000 students
    Online now
  • Hero
    College Math Guru
    Online now

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

aceace

hey can you guys please help with this question...

  • one year ago
  • one year ago

  • This Question is Closed
  1. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    cmn fast

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

    \[1/2\log_{2}x=1/3 \log_{2}y -1\] can you please simplfy

    • one year ago
  3. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    \[\log_{2}{x^2}=\log_{2}{(y-1)^3} \]now it is clear that \[x^2=(y-1)^3\]so now solve the equation & gt ur answer:)

    • one year ago
  4. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    and i use the property\[\frac{1}{e}\log_{b}{a}=\log_{b}{a^e} \]

    • one year ago
  5. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    can u do it from here @aceace ????

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

    i dont really get the formula thing and i dont think i can do it...

    • one year ago
  7. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    k! i give u the solution

    • one year ago
  8. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    using the identity\[(x-y)^3=x^3-y^3-3x^2y-3xy^2\]

    • one year ago
  9. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    so,\[x^2=y^3-1^3-3y^21-3y1^2\]

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

    i think the -1 may be separate because that is the wrong answer...

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

    @jiteshmeghwal9 formula should be: \[\large e.Log_ba = Log_b(a)^e\]

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

    what is that formula for?

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

    This is known as Power rule in Logarithms.. For example: \[Log(x - 1)^2 \implies 2.Log(x-1)\] Or, \[2.Log(x-1) \implies Log(x-1)^2\]

    • one year ago
  14. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    i meant if \[\LARGE{\log_{a^e}{b}=1/e \log_{a}{b}}\]then it will be\[\LARGE{\log_{a}{b^e} }\]

    • one year ago
  15. jiteshmeghwal9
    Best Response
    You've already chosen the best response.
    Medals 1

    so i m correct

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

    you are asked to simplify \[\frac{1}{2}\log _{2}(x) = \frac{1}{3}\log _{2}(y -1)\] multiply both sides by 3 \[\log_{2}(y -1) = \frac{3}{2}\log(x)\]

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

    ok.... that doesn't solve my problem...

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

    now multiply both sides by 2 \[2\log _{2}(y -1) = 3\log_{2}(x)\]

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

    i think that the -1 is separate from teh log on the right hand side...

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

    that is wrong

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

    \[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[\large \log_2(x)^\frac{1}{2} + Log_22 = Log_2(y)^\frac{1}{3}\] \[\large Log_2(2\sqrt{x}) = Log_2(\sqrt[3]{y})\] \[\Large 2 \times \sqrt{x} = \sqrt[3]{y}\]

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

    that is what i got @waterineyes but the answer was different

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

    What is the answer tell me??

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

    it was...\[64x^{3}=y^{2}\]

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

    @waterineyes

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

    Yeah I got it...

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

    Before doing that I want to tell that: \[largeLog_2 64 = 6\]

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

    \[\LARGE (2\sqrt{x})^6 = 64x^{3}\] \[\LARGE (\sqrt[3]{y})^6 = y^2\] The answer you got IS correct. Just a different form.

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

    \[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[3\log_2x + 6= 2\log_2y\] \[\large \log_2x^3 + Log_264= \log_2y^2\] \[64x^3 = y^2\]

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

    here is the solution \[\frac{1}{2}\log_{2}(x) + log_{2}(2) = \frac{1}{3}\log_{2}(y)\] multiply every term by 6 \[3\log_{2}(x) + 6\log_{2}(2) = 2\log_{2}(y)\] then \[Log_{2}(2^6x^3) = y^2\] raise to the power of 2 \[64x^3 = y^2\]

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

    what is large LOG?

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

    what does it mean?

    • one year ago
  33. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

    here is a constant source of confusion since log is a function it is really best to write \(\log(x)\) rather than \(\log x\)

    • one year ago
  34. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

    many (in fact maybe most) texts do not do this, but it would clear up the difference, for example, between \(\log(x+1)\)and \(\log (x) + 1\) then gets obscured when you simply write \(\log x + 1\)

    • one year ago
  35. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

    i am assuming the problem is \[\frac{1}{2}\log(x)=\frac{1}{3}\log(y-1)\] but it is hard to know from the way it is written

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

    ok but what is a large log? as seen above

    • one year ago
  37. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

    who knows?

    • one year ago
  38. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

    whatever it is, if the problem is the one i wrote you can multiply both sides by 6 and get \[3\log(x)=2\log(y-1)\] \[\log(x^3)=\log((y-1)^2)\] and so \[x^3=(y-1)^2\]

    • one year ago
  39. satellite73
    Best Response
    You've already chosen the best response.
    Medals 0

    if was something else, say \[\frac{1}{2}\log(x)=\frac{1}{3}\log(y) - 1\] then proceed as @campbell_st above

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

    @aceace Large Log isn't a mathematical term. @waterineyes was trying to make the text bigger using a LaTeX markup command. Looks like a slash was missing, that's all.

    • one year ago
    • Attachments:

See more questions >>>

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.