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aceace

  • 2 years ago

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

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  1. jiteshmeghwal9
    • 2 years ago
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    cmn fast

  2. aceace
    • 2 years ago
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    \[1/2\log_{2}x=1/3 \log_{2}y -1\] can you please simplfy

  3. jiteshmeghwal9
    • 2 years ago
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    \[\log_{2}{x^2}=\log_{2}{(y-1)^3} \]now it is clear that \[x^2=(y-1)^3\]so now solve the equation & gt ur answer:)

  4. jiteshmeghwal9
    • 2 years ago
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    and i use the property\[\frac{1}{e}\log_{b}{a}=\log_{b}{a^e} \]

  5. jiteshmeghwal9
    • 2 years ago
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    can u do it from here @aceace ????

  6. aceace
    • 2 years ago
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    i dont really get the formula thing and i dont think i can do it...

  7. jiteshmeghwal9
    • 2 years ago
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    k! i give u the solution

  8. jiteshmeghwal9
    • 2 years ago
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    using the identity\[(x-y)^3=x^3-y^3-3x^2y-3xy^2\]

  9. jiteshmeghwal9
    • 2 years ago
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    so,\[x^2=y^3-1^3-3y^21-3y1^2\]

  10. aceace
    • 2 years ago
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    i think the -1 may be separate because that is the wrong answer...

  11. waterineyes
    • 2 years ago
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    @jiteshmeghwal9 formula should be: \[\large e.Log_ba = Log_b(a)^e\]

  12. aceace
    • 2 years ago
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    what is that formula for?

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

  14. jiteshmeghwal9
    • 2 years ago
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    i meant if \[\LARGE{\log_{a^e}{b}=1/e \log_{a}{b}}\]then it will be\[\LARGE{\log_{a}{b^e} }\]

  15. jiteshmeghwal9
    • 2 years ago
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    so i m correct

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

  17. aceace
    • 2 years ago
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    ok.... that doesn't solve my problem...

  18. campbell_st
    • 2 years ago
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    now multiply both sides by 2 \[2\log _{2}(y -1) = 3\log_{2}(x)\]

  19. aceace
    • 2 years ago
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    i think that the -1 is separate from teh log on the right hand side...

  20. aceace
    • 2 years ago
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    that is wrong

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

  22. aceace
    • 2 years ago
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    that is what i got @waterineyes but the answer was different

  23. waterineyes
    • 2 years ago
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    What is the answer tell me??

  24. aceace
    • 2 years ago
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    it was...\[64x^{3}=y^{2}\]

  25. aceace
    • 2 years ago
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    @waterineyes

  26. waterineyes
    • 2 years ago
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    Yeah I got it...

  27. waterineyes
    • 2 years ago
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    Before doing that I want to tell that: \[largeLog_2 64 = 6\]

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

  29. waterineyes
    • 2 years ago
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    \[\frac{1}{2}\log_2x=\frac{1}{3}\log_2y−1\] \[\frac{1}{2}\log_2x + 1=\frac{1}{3}\log_2y\] \[3\log_2x + 6= 2\log_2y\] \[\large \log_2x^3 + Log_264= \log_2y^2\] \[64x^3 = y^2\]

  30. campbell_st
    • 2 years ago
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    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 \[64x^3 = y^2\]

  31. aceace
    • 2 years ago
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    what is large LOG?

  32. aceace
    • 2 years ago
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    what does it mean?

  33. satellite73
    • 2 years ago
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    here is a constant source of confusion since log is a function it is really best to write \(\log(x)\) rather than \(\log x\)

  34. satellite73
    • 2 years ago
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    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 \(\log x + 1\)

  35. satellite73
    • 2 years ago
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    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

  36. aceace
    • 2 years ago
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    ok but what is a large log? as seen above

  37. satellite73
    • 2 years ago
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    who knows?

  38. satellite73
    • 2 years ago
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    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 \[x^3=(y-1)^2\]

  39. satellite73
    • 2 years ago
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    if was something else, say \[\frac{1}{2}\log(x)=\frac{1}{3}\log(y) - 1\] then proceed as @campbell_st above

  40. Wired
    • 2 years ago
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    @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.

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