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March 30, 2017

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still having a hard time understand these? please help?

Logx10 + logx8-logx5=4

Log3256-log327=x

Log 2+ 3log 5-log 25=x

  • math - ,

    add logs to multiply, subtract to divide
    logx (10*8/5) =4

    logx (16) = 4

    in general base^ logx = x
    so
    16 = x^4
    x = 16^(1/4) = (2*2*2*2)^1/4 = 2

  • math - ,

    Log3 256-log3 27=x
    same as last question, subtract is divide
    log3 (256/27) = x
    256/27 = 3^x
    I bet you have a typo because 27 is 3^3 but 256 is nothing to the 3 so have to use calculator
    anyway
    9.48 = 3^x
    log10 (9.48) = x log10 (3)
    .9768 = x * .4771
    x = .9768/.4771

  • math - ,

    base 10 I assume

    Log 2+ 3log 5-log 25=x

    log 2 + log 125 - log 25 = x

    log (250/25) = x

    log 10 = 1 = x

  • math - ,

    ok so by your previous help i tried to solve this problem
    Log32 56-log32 7=x (log base 32)
    what i did was 57/7=8
    so in log form i got log 32 8=x ( log base 32)
    then changed it to exponential form to 32^x=8
    x= 0.6
    correct?

  • math - ,

    yes i do have a typo
    Log32 56-log32 7=x
    and my work is above

  • math - ,

    log 32 (8) = x

    8 = 32^x

    2^3 = 2^5x

    x = 3/5 = 6/10 = 0.6 yes

    I suspected you did not need a calculator :)

  • math - ,

    It helps if things like 2^3 = 8 and 2^5 = 32 are part of your tool box.

  • math - ,

    also very common
    3^3 = 27
    5^3 = 125

  • math - ,

    sorry to bug you but i think this may be the last problem
    2log4 x=3 (log base 4)
    i got log16^x=3
    16^3=x
    x=4096
    right?

  • math - ,

    log4 (x^2) = 3
    x^2 = 4^3 = 64
    x = 8

  • math - ,

    log 32 x = 6/10

  • math - ,

    log 625 3/4 = x

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