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Log equations

  1. Apr 23, 2007 #1
    1. The problem statement, all variables and given/known data
    Solve Algebraically:

    Log(subscript4) (a^2 + 2) = Log(subscript 4) (2a + 10)


    2. Relevant equations
    other equations that are similar which I can't do are:
    Solve Algebraically:

    -Log(subscript 5) (x + 3) + Log(subscript 5) (x - 2) = Log(subscript 5) 14

    -Log(subscript 2) (2x + 6) - Log(subscript 2) x = 3




    3. The attempt at a solution
    not sure if this is right but I think you would divide by Log(subscript 4)
    Log(subscript4) (a^2 + 2) = Log(subscript 4) (2a + 10)
    ------------------------------------------------------
    Log(subscript4) Log(subscript4)

    canceled those out leaving:
    a^2 + 2 = 2a + 10

    expand the equation:
    this is where i get lost, unless im doing it completely wrong. Please help.
     
  2. jcsd
  3. Apr 23, 2007 #2

    Dick

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    Your solution is essentially right, but the formalism is painfully wrong. If f(x)=f(y) you cannot divide by f. You have to apply the inverse function of f. The inverse of Log(subscript4) is 4^. As in 4^(Log(subscript4)(x))=x. And a^2 + 2 = 2a + 10
    is a harmless quadratic equation. Don't get confused there, just factor it or use the quadratic equation.
     
  4. Apr 23, 2007 #3

    VietDao29

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    When doing log equations, you should try to remember all the formulae. Some useful are:

    (1) logab + logac = loga(bc)
    (2) logab - logac = loga(b / c)
    (3) loga(bc) = c loga(b)
    (4) The change base formula:
    [tex]\log_{a}b = \frac{\log_{c}b}{\log_{c}a}[/tex]

    (5) Since logarithm is either strictly increasing or strictly decreasing (i.e, 1-1 function), we have:
    [tex]\log_a b = \log_a c \Leftrightarrow b = c[/tex]

    You cannot divide logab by loga to get b, it's as meaningless as saying that:
    [tex]\frac{\sin x}{\sin} = x[/tex]

    ----------------

    I'll help you do the first one, and let you try the other 2 by yourself. See if you can get it. :)

    log4 (a2 + 2) = log4 (2a + 10)
    Using (5), we have:
    <=> a2 + 2 = 2a + 10
    <=> a2 - 2a - 8 = 0
    [tex]\Leftrightarrow a = 4 \quad \mbox{or} \quad a = -2[/tex]
    a2 + 2, and 2a + 10 should all be positive. And both values satisfy the requirements, so we have 2 solutions:
    [tex]a = 4 \quad \mbox{or} \quad a = -2[/tex]

    Is it cear?
    Can you do the other 2? :)
     
    Last edited: Apr 23, 2007
  5. Apr 23, 2007 #4

    AKG

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    VietDao, why should a be positive? You need a2+2 and 2a+10 to be positive, you don't need a to be positive. Note, those expressions are positive for both a=4 and a=-2.
     
  6. Apr 23, 2007 #5

    VietDao29

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    Whoops, I thought that a was the base :blushing: :cry:. Er... stand corrected. :smile:
    Thanks. :blushing:
     
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