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anonymous

  • one year ago

49log2*2log7 =? please do a favour to solve this ques.

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  1. rajat97
    • one year ago
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    it is in the form xlogy*ylogx

  2. rajat97
    • one year ago
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    what can you do to simplify it?

  3. Loser66
    • one year ago
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    .

  4. anonymous
    • one year ago
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    |dw:1436709166375:dw|

  5. anonymous
    • one year ago
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    can i do somthing like that

  6. Loser66
    • one year ago
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    I don't think you can do anything with the original one unless using calculator. It is simplest form since (2,7) are relative prime.

  7. anonymous
    • one year ago
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    oke

  8. anonymous
    • one year ago
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    49log2*2log7=7^2log2*2log7

  9. anonymous
    • one year ago
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    what wud be the next step.

  10. SolomonZelman
    • one year ago
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    you are multiplying logarithms (correct?) \(\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) }\) At first I will ask you, are you sure that it is supposed to be a "•", and not a "+" ?

  11. anonymous
    • one year ago
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    yes it is sure http://www.m4maths.com/placement-puzzles.php?SOURCE=amcat&TOPIC=Numerical%20Ability&SUB_TOPIC=Ratio%20and%20Proportion you can check out at this link

  12. SolomonZelman
    • one year ago
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    I don't have a login to this thing, but I of course believe you. I was just asking that for a clarification.....

  13. anonymous
    • one year ago
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    oke but i wanna ask is there any formula to solve the above equation , which i hve done by drawing. log 2^7^2

  14. SolomonZelman
    • one year ago
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    I think that you can't make it any simpler than \(\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) =98\log(2)\log(7)}\) (And of course the assumed base is 10, not e)

  15. anonymous
    • one year ago
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    yes!

  16. SolomonZelman
    • one year ago
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    Assumed base is also not important tho' for this multiplication, because the assumed base wouldn't matter with this multiplication.

  17. SolomonZelman
    • one year ago
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    All we did, is kinda: \(\large\color{black}{ \displaystyle 3a\cdot 4x=12ax }\)

  18. SolomonZelman
    • one year ago
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    we multiplied the coefficients, but you can't really put the logarithms together more than "log(2)log(7)"

  19. anonymous
    • one year ago
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    yaa

  20. SolomonZelman
    • one year ago
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    questions about this?

  21. anonymous
    • one year ago
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    so we need to put the values of log 2 and log 7 to calculate the exact answer?

  22. anonymous
    • one year ago
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    as we cannt do anthing with them further

  23. SolomonZelman
    • one year ago
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    If you calculate these values it would be an approximation, and the "exact answer" is what we wrote before 98log(2)log(7)

  24. anonymous
    • one year ago
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    49log2*2log7=7^2log2*2log7=7log2^2*2log7=7*2log7=2 this is the solution by somebody on another site . pls see to it

  25. SolomonZelman
    • one year ago
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    No, it is not equivalent to 2. THAT IS WRONG....

  26. anonymous
    • one year ago
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    i didnt get the logic here

  27. anonymous
    • one year ago
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    yaa

  28. SolomonZelman
    • one year ago
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    no logic... it is wrong... if it had logic it would be much more frightening:O

  29. anonymous
    • one year ago
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    i have solved it , we have to take 2 as a base in frst case

  30. SolomonZelman
    • one year ago
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    I can't really follow what you did....

  31. anonymous
    • one year ago
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    i have dun blunder to this ques *O*

  32. anonymous
    • one year ago
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    hahaha

  33. SolomonZelman
    • one year ago
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    we can't take 2 as the base if it is a log.... you can't rewrite what's in the log, as the base... I mean you can't, but nothing correct is going to get out of it.

  34. SolomonZelman
    • one year ago
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    As again, the only thing here that I see valid is: \(\large\color{black}{ \displaystyle 49\log(2)~\cdot~2\log(7) =98\log(2)\log(7)}\)

  35. anonymous
    • one year ago
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    ok

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