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Algebra 2

posted by on .

how do not understand how to do this

Log X + Lob (X-3) = 1

I know I do this

Log X(X-3)=1

then I do this

Log X^(2) - 3X = 1

then I do this

2 Log (X-3X) = 1

then

2 Log (-2X) = 1

then

(2 Log (-2X) = 1)(1/2)

Log (-2X) = 1/2

then

(Log (-2X) = 1/2)(-1/2)

Log X = (-1/4)

then

by def

10^ (-1/4) = X

I get something under one and calculator tells me I can't do that I think I'm doing something wrong what do I do?

  • Algebra 2 - ,

    Log X + Lob (X-3) = 1

    I know I do this

    Log X(X-3)=1

    Then take the antilog of each side
    x(x-3)=0
    Now solve it.

  • Algebra 2 - ,

    what exactly is antilog?

  • Algebra 2 - ,

    your notation at Log X^(2) - 3X = 1 is sloppy

    say: Log (X^(2) - 3X) = 1

    your next line of 2 Log (X-3X) = 1
    is WRONG

    from Log (X^(2) - 3X) = 1 by definition of logs
    x^2 - 3x = 10^1
    x^2 - 3x - 10 = 0
    (x-5)(x+2) = 0
    x = 5 or x = -2

    but in logx, x > 0, so

    x = 5

    check: if x=5
    LS = log5 + log2
    = log(5x2)
    = log 10
    = 1
    = RS

  • Algebra 2 - ,

    ANSWER=3

  • Algebra 2 - ,

    ok um so
    Log (X^(2) - 3X) = 1
    x^2 - 3x = 10^1

    what allows you to put in the value ten and get ride of the log on the other side? What if the value wasn't one would u just do 10^X?

    x^2 - 3x - 10 = 0
    (x-5)(x+2) = 0
    x = 5 or x = -2

  • Algebra 2 - ,

    NO, it is x=5
    I verified it

    Here is yours:
    if x=3
    LS= log3 + log1
    = log(3x1)
    = log 3 which is not equal to the RS of 1

  • Algebra 2 - ,

    LOGb<?sub> a = c <----> bc = a

    e.g.
    log2 8 = 3 <---> 23 = 8

  • Algebra 2 - ,

    let's try that again

    LOGb a = c <----> bc = a

    e.g.
    log2 8 = 3 <---> 23 = 8

  • Algebra 2 - ,

    ok um so
    Log (X^(2) - 3X) = 1
    x^2 - 3x = 10^1

    what allows you to put in the value ten and get ride of the log on the other side? What if the value wasn't one would u just do 10^X?

    x^2 - 3x - 10 = 0
    (x-5)(x+2) = 0
    x = 5 or x = -2

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