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Whats the answer to dis one?helppppppppppppp

  1. Apr 26, 2005 #1
    whats the answer to dis one???helppppppppppppp!!!

    3log_6 - log_6 12 + log_6 2 = ???? (the base is 6)

    whats the vertical asymptote, x-intercept, range and domain for this one -
    log_x 4
  2. jcsd
  3. Apr 26, 2005 #2


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    Use the basic properties of logarithms:

    log_b(xy) = log_b(x) + log_b(y)
    log_b(x^y) = y log_b(x)
    log_b(x/y) = log_b(x) - log_b(y)
  4. Apr 26, 2005 #3


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    As for the second,use

    [tex] \log_{4}x=\frac{\ln x}{\ln 4} [/tex]

  5. Apr 27, 2005 #4

    how did u get dat?
    logx4 = In x/In 4??
    Last edited: Apr 28, 2005
  6. Apr 27, 2005 #5


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    Saying "y= log4 x" is the same as saying "x= 4y". Now take the logarithm (with whatever base you want- log10 or ln if you want to use a calculator): log x= log 4y[/sup= y log 4 so y= log x/log 4.
  7. Apr 28, 2005 #6

    x is the base not 4, logx4,, how oh how do u?
    Last edited: Apr 28, 2005
  8. Apr 28, 2005 #7


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    What do you want to do with logx 4? If you x is a given number, then you would find logx4, just as before: (log 4)/(log x). Again, the log can be either base 10 or natural log whichever is easier.
  9. Apr 28, 2005 #8
    yooooooooooo help

    find the domain, range ang x-intercept y=logx4
    Last edited: Apr 29, 2005
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