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Math Help - Log Change of Base Solution Please

  1. #1
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    Log Change of Base Solution Please

    My daughter has the following question on her home work. Although I'm pretty good on all the calculus for the life of me I can not figure this out and am obviously doing some thing wrong. If you can help that would be great.

    The question is:

    Using the log change of base rule solve for x:

    Log 56x+4 = Log 2x49

    Should be simple...right?

    Thanks
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  2. #2
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    Re: Log Change of Base Solution Please

    Hey tommly.

    The change of base rule says that log_a(x) = log_b(x)/log_b(a) for bases a and b.

    Hint: Try changing the LHS to log_2(x) given that it is log_5(x) by using this formula.
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  3. #3
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    Re: Log Change of Base Solution Please

    Quote Originally Posted by tommiy View Post
    My daughter has the following question on her home work. Although I'm pretty good on all the calculus for the life of me I can not figure this out and am obviously doing some thing wrong. If you can help that would be great.

    The question is:

    Using the log change of base rule solve for x:

    Log 56x+4 = Log 2x49

    Should be simple...right?

    Thanks
    Is this the equation? It's a bit hard to read. \displaystyle \log_5{(6x+4)} = \log_{2x}{(49)}
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  4. #4
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    Re: Log Change of Base Solution Please

    doing the Left hand side I then get

    Log 2x (6x+4)/Log 2x 5 = Log 2x49


    That does not really help me.

    I've tried changing the RHS to Log 5 and that does not help me either. I end up with a Log multiplying a Log both the the variable x in it......

    obviously missing something
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  5. #5
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    Re: Log Change of Base Solution Please

    guess no one knows .... i'll continue to try
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  6. #6
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    Re: Log Change of Base Solution Please

    Hello, tommiy!

    Your use of [SUB] and [SUP] is confusing.

    \begin{array}{ccc}\text{It could mean: }& \log_5(6x+4) \:=\:\log_{2x}49 & [1] \\ \\[-3mm] \text{Or maybe: }& \log(5^{6x+4}) \:=\:\log(2x^{49}) & [2] \end{array}


    If it is [1], we have: . \log_5(6x+4) \:=\:\frac{\log_5(49)}{\log_5(2x)}

    . . which becomes: . \log_5(2x)\cdot\log_5(6x+4) \:=\:\log_5(49)

    And I see no way to solve for x.


    If it is [2], we have: . (6x+4)\log5 \:=\:\log(2x^{49})

    This is a transcendental equation.
    It cannot be solved for x.


    From what course did this problem arise?
    And who would assign such a frustrating problem?
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  7. #7
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    Re: Log Change of Base Solution Please

    Its number 1 and its from there standard text book for Year 12 in Australia. It actually has he answer of 3.5 which is correct but I can not see how you manage to achieve this.
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