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yeller33 Group Title

logx-log6=log15

  • 2 years ago
  • 2 years ago

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  1. sauravshakya Group Title
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    HINT: logx-log6 = log(x/6)

    • 2 years ago
  2. sauravshakya Group Title
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    Now, can u?

    • 2 years ago
  3. TheViper Group Title
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    so \(\frac{x}{6}=15\) ?? @sauravshakya

    • 2 years ago
  4. sauravshakya Group Title
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    Yep... @TheViper

    • 2 years ago
  5. TheViper Group Title
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    Yeah I hoped :) THANX ;)

    • 2 years ago
  6. sauravshakya Group Title
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    WElcome

    • 2 years ago
  7. vishweshshrimali5 Group Title
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    @yeller33 see, log a- log b = log(a/b) and log a + log b = log (ab) remember these 2 formula for ever

    • 2 years ago
  8. abayomi12 Group Title
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    log(x) - log(6) = log(15) Add log(6) to each side: log(x) = log(15) + log(6) = log(15 times 6) x = 15 times 6 x = 90

    • 2 years ago
  9. vishweshshrimali5 Group Title
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    Gud

    • 2 years ago
  10. sauravshakya Group Title
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    |dw:1345303238963:dw|And , right? @vishweshshrimali5

    • 2 years ago
  11. vishweshshrimali5 Group Title
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    Yes @sauravshakya absolutely correct

    • 2 years ago
  12. yeller33 Group Title
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    okay i understand thanks guys!

    • 2 years ago
  13. sauravshakya Group Title
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    Welcome

    • 2 years ago
  14. yeller33 Group Title
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    how do you do it if it has a number at the end. = log7+logx=2.... the 2 throws me off.

    • 2 years ago
  15. Kaederfds Group Title
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    x is 90

    • 2 years ago
  16. sauravshakya Group Title
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    log7 + logx=2 log(7*x)=2

    • 2 years ago
  17. sauravshakya Group Title
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    Does it help?

    • 2 years ago
  18. vishweshshrimali5 Group Title
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    and remember that \[\large log_a a = 1\]

    • 2 years ago
  19. yeller33 Group Title
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    x=2//7?? ( 7x=2 )

    • 2 years ago
  20. vishweshshrimali5 Group Title
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    Thus u can write 2 = 2* 1 = \(\large 2* log_{10} 10\) also

    • 2 years ago
  21. sauravshakya Group Title
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    log7 + logx=2 log(7*x)=2 10^2 = 7*x 100/7=x

    • 2 years ago
  22. vishweshshrimali5 Group Title
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    \[\large b\space log _a x = log_a (b^x)\]

    • 2 years ago
  23. vishweshshrimali5 Group Title
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    That's how @sauravshakya got the third step

    • 2 years ago
  24. sauravshakya Group Title
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    got it? @yeller33

    • 2 years ago
  25. vishweshshrimali5 Group Title
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    i can write 2 = \(\large \log_{10} (10^2)\)

    • 2 years ago
  26. sauravshakya Group Title
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    Yep @vishweshshrimali5

    • 2 years ago
  27. vishweshshrimali5 Group Title
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    :)

    • 2 years ago
  28. sauravshakya Group Title
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    log7 + logx=2 log(7*x)=2 1/2 log(7x)=1 log{(7x)^1/2} = log10 (7x)^(1/2) = 10 7x = 100 x=100/7

    • 2 years ago
  29. sauravshakya Group Title
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    according to @vishweshshrimali5

    • 2 years ago
  30. yeller33 Group Title
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    can you two see if i did this one correctly: logx+log8=1 log8*logx=1 8x=1 8x=10^1 10/8

    • 2 years ago
  31. yeller33 Group Title
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    i tried to follow your explanations.

    • 2 years ago
  32. sauravshakya Group Title
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    dont think so

    • 2 years ago
  33. sauravshakya Group Title
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    your second step is wrong

    • 2 years ago
  34. sauravshakya Group Title
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    logx+log8= log8x

    • 2 years ago
  35. sauravshakya Group Title
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    and 1=log10

    • 2 years ago
  36. sauravshakya Group Title
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    got your mistake?

    • 2 years ago
  37. yeller33 Group Title
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    i think so . thanks again!

    • 2 years ago
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