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mathmath333

  • one year ago

Frustrating problem

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  1. mathmath333
    • one year ago
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    \(\normalsize \color{black}{\begin{align}& \text{if}\ \log 242=a,\log 80=b\ \text{and}\ \log 45=c & \hspace{.33em}\\~\\ & \text{express}\ \log36\ \text{in terms of }\ a,b,\ \text{and }\ c.\hspace{.33em}\\~\\ & a.)\ \dfrac{(c-1)(3c+b-4)}{2} \hspace{.33em}\\~\\ & b.)\ \dfrac{(c-1)(3c+b-4)}{3} \hspace{.33em}\\~\\ & c.)\ \dfrac{(c-1)(3c-b-4)}{2} \hspace{.33em}\\~\\ & (d.)\ \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

  2. phi
    • one year ago
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    one thought is to first find the prime factors of each of the given numbers

  3. mathmath333
    • one year ago
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    i tried and ended up blundering every time in calculation

  4. mathmath333
    • one year ago
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    further i need to solve these problem within 2-3 min

  5. phi
    • one year ago
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    are you allowed a calculator?

  6. mathmath333
    • one year ago
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    lol no / cheating prohibited

  7. imqwerty
    • one year ago
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    try out the options lol

  8. mathmath333
    • one year ago
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    of cos

  9. imqwerty
    • one year ago
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    so the answer is.......

  10. mathmath333
    • one year ago
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    thats a liitle time tough

  11. mathmath333
    • one year ago
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    lol and the book gives none of these at last when i checked

  12. imqwerty
    • one year ago
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    lol xD i hate when it happens xD

  13. mathmath333
    • one year ago
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    the objective here is to i need to observe this problem deeply

  14. imqwerty
    • one year ago
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    nd i m the guy who likes to do the opposite of what is said to be done. XD

  15. mathmath333
    • one year ago
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    what is the minimum time this problem should be solved

  16. imqwerty
    • one year ago
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    1-1.5min

  17. mathmath333
    • one year ago
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    i guess there should be some trick

  18. imqwerty
    • one year ago
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    there is :)

  19. ChillOut
    • one year ago
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    Is the anwser d?

  20. Lurker
    • one year ago
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    |dw:1438120503037:dw|

  21. ChillOut
    • one year ago
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    Please note that: 242= 11²*2, 80=8*10 and 45=90/2

  22. phi
    • one year ago
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    I don't see a fast way. but we can look at the first choice and say log(45/10) ( log(45^3) + log(80)-log(10000)) = log(36^2) or log(45/10) ( log ( 5^3 * 3^6 * 2^3 / (2^2 5^2) ) = ... log(4.5) log(2*5*3^6) = log(2^4 3^4)

  23. mathmath333
    • one year ago
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    i guess this is the only way

  24. ChillOut
    • one year ago
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    I can't find a fast way either.

  25. phi
    • one year ago
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    log (9/2) is about 0.95-0.3 = 0.65 or about 2/3 so log(2^(2/3) 5^(2/3) * 3^4) = log(2^4 * 3^4) log(10^(2/3) ) + log(3^4) = log(2^4) + log(3^4) 2/3 + log(3^4) = log(2^4) + log(3^4) 2/3 = log(2^4) ? definitely not

  26. ChillOut
    • one year ago
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    b=3log2+log10 = 3log2 +1 b-1=3log2

  27. ChillOut
    • one year ago
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    c=log45 -> log (90/2)=log90 - log2 c=2log 3+1-log2 c-1=2log3-log2

  28. ChillOut
    • one year ago
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    b-1=3log2 (b-1)/3=log2 c-1=2log3-(b-1)/3

  29. imqwerty
    • one year ago
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  30. ChillOut
    • one year ago
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    That's it! This is all about getting the smallest common divisors in terms of log(10)

  31. imqwerty
    • one year ago
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    but still i think that tryin out options will be the best

  32. phi
    • one year ago
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    to do that in 2 minutes you will have to get *very* familiar with logs

  33. Loser66
    • one year ago
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    To me, it is "none of them"

  34. mathmath333
    • one year ago
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    i nned to remember log 2 and log 3 values and then try out with options simpifying it only log 2 and log 3

  35. phi
    • one year ago
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    yes , none of the above. But how do you know that quickly?

  36. Loser66
    • one year ago
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    log (80) = log (20*4) = log (20) +log(4) hence log (4) = log(80) - log (20) = b- log(20)

  37. Loser66
    • one year ago
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    36 = 4*9 log 36 = log 4 + log9 = b-log 20 + log9 and log 45 = log 9+ log 5--> log 9 = log 45-log5 = c - log5 hence log36 = b-log20+c-log5 log 36 = b+c-(log20+log5) = b+c-log(20*5) = b+c - log 100 = b+c-2 None of the above can simplify to get b+c-2.

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